THE EQUAL VARIABLE METHOD VASILE CÎRTOAJE Department of Automation and Computers University of Ploieşti Bucureşti 39, Romania. EMail: vcirtoaje@upg-ploiesti.ro Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page Received: 01 March, 2006 Accepted: 17 April, 2006 Communicated by: Contents JJ II P.S. Bullen J I 2000 AMS Sub. Class.: 26D10, 26D20. Page 1 of 41 Key words: Symmetric inequalities, Power means, EV-Theorem. Abstract: The Equal Variable Method (called also n − 1 Equal Variable Method on the Mathlinks Site - Inequalities Forum) can be used to prove some difficult symmetric inequalities involving either three power means or, more general, two power means and an expression of form f (x1 ) + f (x2 ) + · · · + f (xn ). Go Back Full Screen Close Contents 1 Statement of results 3 2 Proofs 8 3 Applications 16 Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page Contents JJ II J I Page 2 of 41 Go Back Full Screen Close 1. Statement of results In order to state and prove the Equal Variable Theorem (EV-Theorem) we require the following lemma and proposition. Lemma 1.1. Let a, b, c be fixed non-negative real numbers, not all equal and at most one of them equal to zero, and let x ≤ y ≤ z be non-negative real numbers such that x + y + z = a + b + c, x p + y p + z p = ap + b p + c p , where p ∈ (−∞, 0] ∪ (1, ∞). For p = 0, the second equation is xyz = abc > 0. Then, there exist two non-negative real numbers x1 and x2 with x1 < x2 such that x ∈ [x1 , x2 ]. Moreover, Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page 1. if x = x1 and p ≤ 0, then 0 < x < y = z; Contents 2. if x = x1 and p > 1, then either 0 = x < y ≤ z or 0 < x < y = z; JJ II 3. if x ∈ (x1 , x2 ), then x < y < z; J I 4. if x = x2 , then x = y < z. Page 3 of 41 Proposition 1.2. Let a, b, c be fixed non-negative real numbers, not all equal and at most one of them equal to zero, and let 0 ≤ x ≤ y ≤ z such that x + y + z = a + b + c, x p + y p + z p = ap + b p + c p , Go Back Full Screen Close where p ∈ (−∞, 0] ∪ (1, ∞). For p = 0, the second equation is xyz =1 abc > 0. Let 0 f (u) be a differentiable function on (0, ∞), such that g(x) = f x p−1 is strictly convex on (0, ∞), and let F3 (x, y, z) = f (x) + f (y) + f (z). 1. If p ≤ 0, then F3 is maximal only for 0 < x = y < z, and is minimal only for 0 < x < y = z; 2. If p > 1 and either f (u) is continuous at u = 0 or lim f (u) = −∞, then F3 u→0 is maximal only for 0 < x = y < z, and is minimal only for either x = 0 or 0 < x < y = z. Theorem 1.3 (Equal Variable Theorem (EV-Theorem)). Let a1 , a2 , . . . , an (n ≥ 3) be fixed non-negative real numbers, and let 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn such that Equal Variable Method Vasile Cîrtoaje x 1 + x 2 + · · · + x n = a1 + a2 + · · · + an , xp1 + xp2 + · · · + xpn = ap1 + ap2 + · · · + apn , vol. 8, iss. 1, art. 15, 2007 where p is a real number, p 6= 1. For p = 0, the second equation is x1 x2 · · · xn = a1 a2 · · · an > 0. Let f (u) be a differentiable function on (0, ∞) such that 1 g(x) = f 0 x p−1 Title Page JJ II is strictly convex on (0, ∞), and let J I Fn (x1 , x2 , . . . , xn ) = f (x1 ) + f (x2 ) + · · · + f (xn ). 1. If p ≤ 0, then Fn is maximal for 0 < x1 = x2 = · · · = xn−1 ≤ xn , and is minimal for 0 < x1 ≤ x2 = x3 = · · · = xn ; 2. If p > 0 and either f (u) is continuous at u = 0 or lim f (u) = −∞, then Fn u→0 is maximal for 0 ≤ x1 = x2 = · · · = xn−1 ≤ xn , and is minimal for either x1 = 0 or 0 < x1 ≤ x2 = x3 = · · · = xn . Remark 1. Let 0 < α <β. If the function f is differentiable on (α, β) and the 1 function g(x) = f 0 x p−1 is strictly convex on (αp−1 , β p−1 ) or (β p−1 , αp−1 ), then the EV-Theorem holds true for x1 , x2 , . . . , xn ∈ (α, β). Contents Page 4 of 41 Go Back Full Screen Close By Theorem 1.3, we easily obtain some particular results, which are very useful in applications. Corollary 1.4. Let a1 , a2 , . . . , an (n ≥ 3) be fixed non-negative numbers, and let 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn such that x 1 + x 2 + · · · + x n = a1 + a2 + · · · + an , x21 + x22 + · · · + x2n = a21 + a22 + · · · + a2n . Equal Variable Method Let f be a differentiable function on (0, ∞) such that g(x) = f 0 (x) is strictly convex on (0, ∞). Moreover, either f (x) is continuous at x = 0 or lim f (x) = −∞. Then, vol. 8, iss. 1, art. 15, 2007 Fn = f (x1 ) + f (x2 ) + · · · + f (xn ) Title Page is maximal for 0 ≤ x1 = x2 = · · · = xn−1 ≤ xn , and is minimal for either x1 = 0 or 0 < x1 ≤ x2 = x3 = · · · = xn . Contents Vasile Cîrtoaje x→0 Corollary 1.5. Let a1 , a2 , . . . , an (n ≥ 3) be fixed positive numbers, and let 0 < x1 ≤ x2 ≤ · · · ≤ xn such that II J I Page 5 of 41 x 1 + x 2 + · · · + x n = a1 + a2 + · · · + an , 1 1 1 1 1 1 + + ··· + = + + ··· + . x1 x2 xn a1 a2 an Let f be a differentiable function on (0, ∞) such that g(x) = f convex on (0, ∞). Then, JJ Go Back Full Screen 0 √1 x is strictly Fn = f (x1 ) + f (x2 ) + · · · + f (xn ) is maximal for 0 < x1 = x2 = · · · = xn−1 ≤ xn , and is minimal for 0 < x1 ≤ x2 = x3 = · · · = xn . Close Corollary 1.6. Let a1 , a2 , . . . , an (n ≥ 3) be fixed positive numbers, and let 0 < x1 ≤ x2 ≤ · · · ≤ xn such that x 1 + x 2 + · · · + x n = a1 + a2 + · · · + an , x 1 x 2 · · · x n = a1 a1 · · · an . Let f be a differentiable function on (0, ∞) such that g(x) = f 0 x1 is strictly convex on (0, ∞). Then, Fn = f (x1 ) + f (x2 ) + · · · + f (xn ) Equal Variable Method is maximal for 0 < x1 = x2 = · · · = xn−1 ≤ xn , and is minimal for 0 < x1 ≤ x2 = x3 = · · · = xn . vol. 8, iss. 1, art. 15, 2007 Corollary 1.7. Let a1 , a2 , . . . , an (n ≥ 3) be fixed non-negative numbers, and let 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn such that Title Page x 1 + x 2 + · · · + x n = a1 + a2 + · · · + an , xp1 + xp2 + · · · + xpn = ap1 + ap2 + · · · + apn , where p is a real number, p 6= 0 and p 6= 1. (a) For p < 0, P = x1 x2 · · · xn is minimal when 0 < x1 = x2 = · · · = xn−1 ≤ xn , and is maximal when 0 < x1 ≤ x2 = x3 = · · · = xn . (b) For p > 0, P = x1 x2 · · · xn is maximal when 0 ≤ x1 = x2 = · · · = xn−1 ≤ xn , and is minimal when either x1 = 0 or 0 < x1 ≤ x2 = x3 = · · · = xn . Corollary 1.8. Let a1 , a2 , . . . , an (n ≥ 3) be fixed non-negative numbers, let 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn such that x 1 + x 2 + · · · + x n = a1 + a2 + · · · + an , xp1 + xp2 + · · · + xpn = ap1 + ap2 + · · · + apn , and let E = xq1 + xq2 + · · · + xqn . Vasile Cîrtoaje Contents JJ II J I Page 6 of 41 Go Back Full Screen Close Case 1. p ≤ 0 (p = 0 yields x1 x2 · · · xn = a1 a2 · · · an > 0). (a) For q ∈ (p, 0) ∪ (1, ∞), E is maximal when 0 < x1 = x2 = · · · = xn−1 ≤ xn , and is minimal when 0 < x1 ≤ x2 = x3 = · · · = xn . (b) For q ∈ (−∞, p)∪(0, 1), E is minimal when 0 < x1 = x2 = · · · = xn−1 ≤ xn , and is maximal when 0 < x1 ≤ x2 = x3 = · · · = xn . Case 2. 0 < p < 1. Equal Variable Method Vasile Cîrtoaje (a) For q ∈ (0, p) ∪ (1, ∞), E is maximal when 0 ≤ x1 = x2 = · · · = xn−1 ≤ xn , and is minimal when either x1 = 0 or 0 < x1 ≤ x2 = x3 = · · · = xn . vol. 8, iss. 1, art. 15, 2007 (b) For q ∈ (−∞, 0)∪(p, 1), E is minimal when 0 ≤ x1 = x2 = · · · = xn−1 ≤ xn , and is maximal when either x1 = 0 or 0 < x1 ≤ x2 = x3 = · · · = xn . Title Page Case 3. p > 1. (a) For q ∈ (0, 1) ∪ (p, ∞), E is maximal when 0 ≤ x1 = x2 = · · · = xn−1 ≤ xn , and is minimal when either x1 = 0 or 0 < x1 ≤ x2 = x3 = · · · = xn . (b) For q ∈ (−∞, 0)∪(1, p), E is minimal when 0 ≤ x1 = x2 = · · · = xn−1 ≤ xn , and is maximal when either x1 = 0 or 0 < x1 ≤ x2 = x3 = · · · = xn . Contents JJ II J I Page 7 of 41 Go Back Full Screen Close 2. Proofs Proof of Lemma 1.1. Let a ≤ b ≤ c. Note that in the excluded cases a = b = c and a = b = 0, there is a single triple (x, y, z) which verifies the conditions x+y+z =a+b+c and xp + y p + z p = ap + bp + cp . Consider now three cases: p = 0, p < 0 and p > 1. A. Case p = 0 (xyz = abc > 0). Let S = by AM-GM Inequality. We have a+b+c 3 x + y + z = 3S, and P = √ 3 Equal Variable Method abc, where S > P > 0 Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 xyz = P 3 , Title Page and from 0 < x ≤ y ≤ z and x < z, it follows that 0 < x < P . Now let √ f = y + z − 2 yz. It is clear that f ≥ 0, with equality if and only if y = z. Writing f as a function of x, r P , f (x) = 3S − x − 2P x we have r P P 0 f (x) = − 1 > 0, x x and hence the function f (x) is strictly increasing. Since f (P ) = 3(S − P ) > 0, the equation f (x) = 0 has a unique positive root x1 , 0 < x1 < P . From f (x) ≥ 0, it follows that x ≥ x1 . Sub-case x = x1 . Since f (x) = f (x1 ) = 0 and f = 0 implies y = z, we have 0 < x < y = z. Contents JJ II J I Page 8 of 41 Go Back Full Screen Close Sub-case x > x1 . We have f (x) > 0 and y < z. Consider now that y and z depend on x. From x + y(x) + z(x) = 3S and x · y(x) · z(x) = P 3 , we get 1 + y 0 + z 0 = 0 0 0 and x1 + yy + zz = 0. Hence, y 0 (x) = y(x − z) , x(z − y) z 0 (x) = z(y − x) . x(z − y) Since y 0 (x) < 0, the function y(x) is strictly decreasing. Since y(x1 ) > x1 (see subcase x = x1 ), there exists x2 > x1 such that y(x2 ) = x2 , y(x) > x for x1 < x < x2 and y(x) < x for x > x2 . Taking into account that y ≥ x, it follows that x1 < x ≤ x2 . On the other hand, we see that z 0 (x) > 0 for x1 < x < x2 . Consequently, the function z(x) is strictly increasing, and hence z(x) > z(x1 ) = y(x1 ) > y(x). Finally, we conclude that x < y < z for x ∈ (x1 , x2 ), and x = y < z for x = x2 . 1 ap +bp +cp p and R = . Taking into account that B. Case p < 0. Denote S = a+b+c 3 3 x + y + z = 3S, xp + y p + z p = 3Rp , 1 Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page Contents JJ II J I from 0 < x ≤ y ≤ z and x < z we get x < S and 3 p R < x < R. Let p −1 y + zp p h = (y + z) − 2. 2 Page 9 of 41 By the AM-GM Inequality, we have Full Screen 1 √ h ≥ 2 yz √ − 2 = 0, yz with equality if and only if y = z. Writing now h as a function of x, p −1 3R − xp p h(x) = (3S − x) − 2, 2 Go Back Close from " # −1−p −p p S R −1 >0 x x 1 it follows that h(x) is strictly increasing. Since h(x) ≥ 0 and h 3 p R = −2, the 3Rp h0 (x) = 2 3Rp − xp 2 1 equation h(x) = 0 has a unique root x1 and x ≥ x1 > 3 p R. Equal Variable Method Sub-case x = x1 . Since f (x) = f (x1 ) = 0, and f = 0 implies y = z, we have 0 < x < y = z. Sub-case x > x1 . We have h(x) > 0 and y < z. Consider now that y and z depend on x. From x+y(x)+z(x) = 3S and xp +y(x)p +z(x)p = 3Rp , we get 1+y 0 +z 0 = 0 and xp−1 + y p−1 y 0 + z p−1 z 0 = 0, and hence y 0 (x) = xp−1 − z p−1 , z p−1 − y p−1 z 0 (x) = xp−1 − y p−1 . y p−1 − z p−1 Since y 0 (x) > 0, the function y(x) is strictly decreasing. Since y(x1 ) > x1 (see subcase x = x1 ), there exists x2 > x1 such that y(x2 ) = x2 , y(x) > x for x1 < x < x2 , and y(x) < x for x > x2 . The condition y ≥ x yields x1 < x ≤ x2 . We see now that z 0 (x) > 0 for x1 < x < x2 . Consequently, the function z(x) is strictly increasing, and hence z(x) > z(x1 ) = y(x1 ) > y(x). Finally, we have x < y < z for x ∈ (x1 , x2 ) and x = y < z for x = x2 . p p p 1 C. Case p > 1. Denoting S = a+b+c and R = a +b3 +c p yields 3 x + y + z = 3S, xp + y p + z p = 3Rp . By Jensen’s inequality applied to the convex function g(u) = up , we have R > S, Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page Contents JJ II J I Page 10 of 41 Go Back Full Screen Close and hence x < S < R. Let 2 h= y+z yp + zp 2 p1 − 1. By Jensen’s Inequality, we get h ≥ 0, with equality if only if y = z. From p 1 2 3R − xp p h(x) = −1 3S − x 2 Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 and 3 h (x) = (3S − x)2 0 3Rp − xp 2 1−p p (Rp − Sxp−1 ) > 0, it follows that the function h(x) is strictly increasing, and h(x) ≥ 0 implies x ≥ x1 . In the case h(0) ≥ 0 we have x1 = 0, and in the case h(0) < 0 we have x1 > 0 and h(x1 ) = 0. Sub-case x = x1 . If h(0) ≥ 0, then 0 = x1 < y(x1 ) ≤ z(x1 ). If h(0) < 0, then h(x1 ) = 0, and since h = 0 implies y = z, we have 0 < x1 < y(x1 ) = z(x1 ). Sub-case x > x1 . Since h(x) is strictly increasing, for x > x1 we have h(x) > h(x1 ) ≥ 0, hence h(x) > 0 and y < z. From x + y(x) + z(x) = 3S and xp + y p (x) + z p (x) = 3Rp , we get y 0 (x) = p−1 p−1 x −z , p−1 z − y p−1 z 0 (x) = p−1 Title Page Contents JJ II J I Page 11 of 41 Go Back Full Screen p−1 y −x . p−1 z − y p−1 Since y 0 (x) < 0, the function y(x) is strictly decreasing. Taking account of y(x1 ) > x1 (see sub-case x = x1 ), there exists x2 > x1 such that y(x2 ) = x2 , y(x) > x for x1 < x < x2 , and y(x) < x for x > x2 . The condition y ≥ x implies x1 < x ≤ x2 . We see now that z 0 (x) > 0 for x1 < x < x2 . Consequently, the function z(x) is Close strictly increasing, and hence z(x) > z(x1 ) ≥ y(x1 ) > y(x). Finally, we conclude that x < y < z for x ∈ (x1 , x2 ), and x = y < z for x = x2 . Proof of Proposition 1.2. Consider the function F (x) = f (x) + f (y(x)) + f (z(x)) defined on x ∈ [x1 , x2 ]. We claim that F (x) is minimal for x = x1 and is maximal for x = x2 . If this assertion is true, then by Lemma 1.1 it follows that: (a) F (x) is minimal for 0 < x = y < z in the case p ≤ 0, or for either x = 0 or 0 < x < y = z in the case p > 1; (b) F (x) is maximal for 0 < x = y < z. In order to prove the claim, assume that x ∈ (x1 , x2 ). By Lemma 1.1, we have 0 < x < y < z. From x + y(x) + z(x) = a + b + c and x + y p (x) + z p (x) = ap + bp + cp , Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page Contents JJ II J I p we get y 0 + z 0 = −1, y p−1 y 0 + z p−1 z 0 = −xp−1 , Page 12 of 41 Go Back Full Screen whence xp−1 − y p−1 xp−1 − z p−1 0 , z = . z p−1 − y p−1 y p−1 − z p−1 It is easy to check that this result is also valid for p = 0. We have y0 = F 0 (x) = f 0 (x) + y 0 f 0 (y) + z 0 f 0 (z) Close and F 0 (x) (xp−1 − y p−1 )(xp−1 − z p−1 ) g(xp−1 ) g(y p−1 ) = p−1 + (x − y p−1 )(xp−1 − z p−1 ) (y p−1 − z p−1 )(y p−1 − xp−1 ) g(z p−1 ) . + p−1 (z − xp−1 )(z p−1 − y p−1 ) Since g is strictly convex, the right hand side is positive. On the other hand, (xp−1 − y p−1 )(xp−1 − z p−1 ) > 0. Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page 0 These results imply F (x) > 0. Consequently, the function F (x) is strictly increasing for x ∈ (x1 , x2 ). Excepting the trivial case when p > 1, x1 = 0 and lim f (u) = −∞, the function F (x) is continuous on [x1 , x2 ], and hence is minimal u→0 only for x = x1 , and is maximal only for x = x2 . Proof of Theorem 1.3. We will consider two cases. Case p ∈ (−∞, 0] ∪ (1, ∞). Excepting the trivial case when p > 1, x1 = 0 and lim f (u) = −∞, the function Fn (x1 , x2 , . . . , xn ) attains its minimum and maximum u→0 values, and the conclusion follows from Proposition 1.2 above, via contradiction. For example, let us consider the case p ≤ 0. In order to prove that Fn is maximal for 0 < x1 = x2 = · · · = xn−1 ≤ xn , we assume, for the sake of contradiction, that Fn attains its maximum at (b1 , b2 , . . . , bn ) with b1 ≤ b2 ≤ · · · ≤ bn and b1 < bn−1 . Let x1 , xn−1 , xn be positive numbers such that x1 + xn−1 + xn = b1 + bn−1 + bn and xp1 + xpn−1 + xpn = bp1 + bpn−1 + bpn . According to Proposition 1.2, the expression F3 (x1 , xn−1 , xn ) = f (x1 ) + f (xn−1 ) + f (xn ) Contents JJ II J I Page 13 of 41 Go Back Full Screen Close is maximal only for x1 = xn−1 < xn , which contradicts the assumption that Fn attains its maximum at (b1 , b2 , . . . , bn ) with b1 < bn−1 . 1 Case p ∈ (0, 1). This case reduces to the case p > 1, replacing each of the ai by aip , 1 each of the xi by xip , and then p by p1 . Thus, we obtain the sufficient condition that 1 0 h(x) = xf x 1−p to be strictly convex on (0, ∞). We claim that this condition is 1 equivalent to the condition that g(x) = f 0 x p−1 to be strictly convex on (0, ∞). Actually, for our proof, it suffices to show that if g(x) is strictly convex on (0, ∞), then h(x) is strictly convex on (0, ∞). To show this, we see that g x1 = x1 h(x). Since g(x) is strictly convex on (0, ∞), by Jensen’s inequality we have u v +y 1 1 x ug + vg > (u + v)g u+v x y for any x, y, u, v > 0 with x 6= y. This inequality is equivalent to ! u v u v u+v h(x) + h(y) > + h u v . x y x y +y x Substituting u = tx and v = (1 − t)y, where t ∈ (0, 1), reduces the inequality to th(x) + (1 − t)h(y) > h(tx + (1 − t)y), which shows us that h(x) is strictly convex on (0, ∞). u→0 p f (u) = , u g(x) = f 0 x 1 p−1 1 = px 1−p , g 00 (x) = Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page Contents JJ II J I Page 14 of 41 Go Back Full Screen Close Proof of Corollary 1.7. We will apply Theorem 1.3 to the function f (u) = p ln u. We see that lim f (u) = −∞ for p > 0, and 0 Equal Variable Method 2p−1 p2 1−p . x (1 − p)2 Since g 00 (x) > 0 for x > 0, the function g(x) is strictly convex on (0, ∞), and the conclusion follows by Theorem 1.3. Proof of Corollary 1.8. We will apply Theorem 1.3 to the function f (u) = q(q − 1)(q − p)uq . For p > 0, it is easy to check that either f (u) is continuous at u = 0 (in the case q > 0) or lim f (u) = −∞ (in the case q < 0). We have u→0 Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 f 0 (u) = q 2 (q − 1)(q − p)uq−1 and Title Page g(x) = f 0 x g 00 (x) = 1 p−1 = q 2 (q − 1)(q − p)x q−1 p−1 , q 2 (q − 1)2 (q − p)2 2p−1 x 1−p . (p − 1)2 Contents JJ II J I 00 Since g (x) > 0 for x > 0, the function g(x) is strictly convex on (0, ∞), and the conclusion follows by Theorem 1.3. Page 15 of 41 Go Back Full Screen Close 3. Applications Proposition 3.1. Let x, y, z be non-negative real numbers such that x + y + z = 2. ln 2 If r0 ≤ r ≤ 3, where r0 = ln 3−ln ≈ 1.71, then 2 xr (y + z) + y r (z + x) + z r (x + y) ≤ 2. Proof. Rewrite the inequality in the homogeneous form r+1 x+y+z r+1 r+1 r+1 x +y +z +2 ≥ (x + y + z)(xr + y r + z r ), 2 and apply Corollary 1.8 (case p = r and q = r + 1): If 0 ≤ x ≤ y ≤ z such that x + y + z = constant and x + y r + z r = constant, r then the sum xr+1 + y r+1 + z r+1 is minimal when either x = 0 or 0 < x ≤ y = z. Case x = 0. The initial inequality becomes yz(y r−1 +z r−1 ) ≤ 2, where y + z = 2. Since 0 < r − 1 ≤ 2, by the Power Mean inequality we have 2 r−1 y r−1 + z r−1 y + z2 2 ≤ . 2 2 Thus, it suffices to show that yz y2 + z2 2 r−1 2 ≤ 1. Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page Contents JJ II J I Page 16 of 41 Go Back Full Screen Close Taking account of 2(y 2 + z 2 ) y2 + z2 = ≥ 1 and 2 (y + z)2 r−1 ≤ 1, 2 we have y2 + z2 2 r−1 2 y2 + z2 1 − yz ≥ 1 − yz 2 4 yz(y 2 + z 2 ) (y + z) − = 16 2 4 (y − z) = ≥ 0. 16 Case 0 < x ≤ y = z. In the homogeneous inequality we may leave aside the constraint x + y + z = 2, and consider y = z = 1, 0 < x ≤ 1. The inequality reduces to x r+1 1+ − xr − x − 1 ≥ 0. 2 r+1 Since 1 + x2 is increasing and xr is decreasing in respect to r, it suffices to consider r = r0 . Let x r0 +1 f (x) = 1 + − xr0 − x − 1. 2 We have r0 + 1 x r0 f 0 (x) = 1+ − r0 xr0 −1 − 1, 2 2 1 00 r0 + 1 x r0 r 0 − 1 f (x) = 1+ − 2−r0 . r0 4 2 x Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page Contents JJ II J I Page 17 of 41 Go Back Full Screen Close Since f 00 (x) is strictly increasing on (0, 1], f 00 (0+ ) = −∞ and r 1 00 r0 + 1 3 0 f (1) = − r0 + 1 r0 4 2 r0 + 1 3 − r0 = − r0 + 1 = > 0, 2 2 there exists x1 ∈ (0, 1) such that f 00 (x1 ) = 0, f 00 (x) < 0 for x ∈ (0, x1 ), and f 00 (x) > 0 for x ∈ (x1 , 1]. Therefore, the function f 0 (x) is strictly decreasing for x ∈ [0, x1 ], and strictly increasing for x ∈ [x1 , 1]. Since r0 r0 + 1 3 r0 − 1 0 0 f (0) = > 0 and f (1) = − 2 = 0, 2 2 2 there exists x2 ∈ (0, x1 ) such that f 0 (x2 ) = 0, f 0 (x) > 0 for x ∈ [0, x2 ), and f 0 (x) < 0 for x ∈ (x2 , 1). Thus, the function f (x) is strictly increasing for x ∈ [0, x2 ], and strictly decreasing for x ∈ [x2 , 1]. Since f (0) = f (1) = 0, it follows that f (x) ≥ 0 for 0 < x ≤ 1, establishing the desired result. For x ≤ y ≤ z, equality occurs when x = 0 and y = z = 1. Moreover, for r = r0 , equality holds again when x = y = z = 1. Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page Contents JJ II J I Page 18 of 41 Go Back Proposition 3.2 ([12]). Let x, y, z be non-negative real numbers such that xy + yz + zx = 3. If 1 < r ≤ 2, then Full Screen xr (y + z) + y r (z + x) + z r (x + y) ≥ 6. Close Proof. Rewrite the inequality in the homogeneous form r r r x (y + z) + y (z + x) + z (x + y) ≥ 6 xy + yz + zx 3 r+1 2 . For convenience, we may leave aside the constraint xy + yz + zx = 3. Using now the constraint x + y + z = 1, the inequality becomes r+1 1 − x2 − y 2 − z 2 2 r r r x (1 − x) + y (1 − y) + z (1 − z) ≥ 6 . 6 To prove it, we will apply Corollary 1.4 to the function f (u) = −ur (1 − u) for 0 ≤ u ≤ 1. We have f 0 (u) = −rur−1 + (r + 1)ur and g(x) = f 0 (x) = −rxr−1 + (r + 1)xr , g 00 (x) = r(r − 1)xr−3 [(r + 1)x + 2 − r]. Since g 00 (x) > 0 for x > 0, g(x) is strictly convex on [0, ∞). According to Corollary 1.4, if 0 ≤ x ≤ y ≤ z such that x + y + z = 1 and x2 + y 2 + z 2 = constant, then the sum f (x) + f (y) + f (z) is maximal for 0 ≤ x = y ≤ z. Thus, we have only to prove the original inequality in the case x = y ≤ z. This means, to prove that 0 < x ≤ 1 ≤ y and x2 + 2xz = 3 implies xr (x + z) + xz r ≥ 3. r r 3−x2 . 2x Let f (x) = x (x + z) + xz − 3, with z = Differentiating the equation x2 + 2xz = 3 yields z 0 = −(x+z) . x Then, f 0 (x) = (r + 1)xr + rxr−1 z + z r + (xr + rxz r−1 )z 0 = (xr−1 − z r−1 )[rx + (r − 1)z] ≤ 0. The function f (x) is strictly decreasing on [0, 1], and hence f (x) ≥ f (1) = 0 for 0 < x ≤ 1. Equality occurs if and only if x = y = z = 1. Proposition 3.3 ([5]). If x1 , x2 , . . . , xn are positive real numbers such that x1 + x2 + · · · + xn = 1 1 1 + + ··· + , x1 x2 xn Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page Contents JJ II J I Page 19 of 41 Go Back Full Screen Close then 1 1 1 + + ··· + ≥ 1. 1 + (n − 1)x1 1 + (n − 1)x2 1 + (n − 1)xn Proof. We have to consider two cases. Case n = 2. The inequality is verified as equality. Case n ≥ 3. Assume that 0 < x1 ≤ x2 ≤ · · · ≤ xn , and then apply Corollary 1.5 to −(n−1) 1 the function f (u) = 1+(n−1)u for u > 0. We have f 0 (u) = [1+(n−1)u] 2 and 1 −(n − 1)x = √ g(x) = f 0 √ 2, x ( x + n − 1) 3(n − 1)2 g 00 (x) = √ √ 4. 2 x ( x + n − 1) Since g 00 (x) > 0, g(x) is strictly convex on (0, ∞). According to Corollary 1.5, if 0 < x1 ≤ x2 ≤ · · · ≤ xn such that x1 + x2 + · · · + xn = constant and 1 1 1 + + ··· + = constant, x1 x2 xn then the sum f (x1 ) + f (x2 ) + · · · + f (xn ) is minimal when 0 < x1 ≤ x2 = x3 = · · · = xn . Thus, we have to prove the inequality 1 n−1 + ≥ 1, 1 + (n − 1)x 1 + (n − 1)y under the constraints 0 < x ≤ 1 ≤ y and x + (n − 1)y = 1 n−1 + . x y Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page Contents JJ II J I Page 20 of 41 Go Back Full Screen Close The last constraint is equivalent to (n − 1)(y − 1) = y(1 − x2 ) . x(1 + y) Since 1 n−1 + −1 1 + (n − 1)x 1 + (n − 1)y 1 1 n−1 n−1 = − + − 1 + (n − 1)x n 1 + (n − 1)y n 2 (n − 1) (y − 1) (n − 1)(1 − x) = − n[1 + (n − 1)x] n[1 + (n − 1)y] (n − 1)(1 − x) (n − 1)y(1 − x2 ) = − , n[1 + (n − 1)x] nx(1 + y)[1 + (n − 1)y] we must show that x(1 + y)[1 + (n − 1)y] ≥ y(1 + x)[1 + (n − 1)x], Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page Contents JJ II J I Page 21 of 41 which reduces to (y − x)[(n − 1)xy − 1] ≥ 0. Go Back Since y − x ≥ 0, we have still to prove that Full Screen (n − 1)xy ≥ 1. Indeed, from x + (n − 1)y = 1 x + n−1 y we get xy = (n − 1)xy − 1 = Close y+(n−1)x , x+(n−1)y and hence n(n − 2)x > 0. x + (n − 1)y For n ≥ 3, one has equality if and only if x1 = x2 = · · · = xn = 1. Proposition 3.4 ([10]). Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. If m is a positive integer satisfying m ≥ n − 1, then 1 1 1 m m m a1 + a2 + · · · + an + (m − 1)n ≥ m + + ··· + . a1 a2 an Proof. For n = 2 (hence m ≥ 1), the inequality reduces to Equal Variable Method m am 1 + a2 + 2m − 2 ≥ m(a1 + a2 ). Vasile Cîrtoaje We can prove it by summing the inequalities ≥ 1 + m(a1 − 1) and ≥ 1+ m(a2 − 1), which are straightforward consequences of Bernoulli’s inequality. For n ≥ 3, replacing a1 , a2 , . . . , an by x11 , x12 , . . . , x1n , respectively, we have to show that vol. 8, iss. 1, art. 15, 2007 1 1 1 + m + · · · + m + (m − 1)n ≥ m(x1 + x2 + · · · + xn ) m x1 x2 xn Contents am 1 am 2 for x1 x2 · · · xn = 1. Assume 0 < x1 ≤ x2 ≤ · · · ≤ xn and apply Corollary 1.8 (case p = 0 and q = −m): If 0 < x1 ≤ x2 ≤ · · · ≤ xn such that x1 + x2 + · · · + xn = constant x1 x2 · · · xn = 1, and then the sum x1m + x1m + · · · + x1m is minimal when 0 < x1 = x2 = · · · = xn−1 ≤ xn . n 1 2 Thus, it suffices to prove the inequality for x1 = x2 = · · · = xn−1 = x ≤ 1, xn = y and xn−1 y = 1, when it reduces to: n−1 1 + m + (m − 1)n ≥ m(n − 1)x + my. m x y Title Page JJ II J I Page 22 of 41 Go Back Full Screen Close By the AM-GM inequality, we have n−1 m + (m − n + 1) ≥ n−1 = my. m x x Then, we have still to show that 1 − 1 ≥ m(n − 1)(x − 1). ym Equal Variable Method Vasile Cîrtoaje This inequality is equivalent to vol. 8, iss. 1, art. 15, 2007 xmn−m − 1 − m(n − 1)(x − 1) ≥ 0 Title Page and (x − 1)[(xmn−m−1 − 1) + (xmn−m−2 − 1) + · · · + (x − 1)] ≥ 0. The last inequality is clearly true. For n = 2 and m = 1, the inequality becomes equality. Otherwise, equality occurs if and only if a1 = a2 = · · · = an = 1. Proposition 3.5 ([6]). Let x1 , x2 , . . . , xn be non-negative real numbers such that x1 + x2 + · · · + xn = n. If k is a positive integer satisfying 2 ≤ k ≤ n + 2, and k−1 n r = n−1 − 1, then Contents JJ II J I Page 23 of 41 Go Back xk1 + xk2 + · · · + xkn − n ≥ nr(1 − x1 x2 · · · xn ). Full Screen Proof. If n = 2, then the inequality reduces to xk1 + xk2 − 2 ≥ (2k − 2)x1 x2 . For k = 2 and k = 3, this inequality becomes equality, while for k = 4 it reduces to 6x1 x2 (1 − x1 x2 ) ≥ 0, which is clearly true. Consider now n ≥ 3 and 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn . Towards proving the inequality, we will apply Corollary 1.7 (case p = k > 0): If 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn Close such that x1 + x2 + · · · + xn = n and xk1 + xk2 + · · · + xkn = constant, then the product x1 x2 · · · xn is minimal when either x1 = 0 or 0 < x1 ≤ x2 = x3 = · · · = xn . Case x1 = 0. The inequality reduces to xk2 + · · · + xkn ≥ nk , (n − 1)k−1 with x2 + · · · + xn = n, This inequality follows by applying Jensen’s inequality to the convex function f (u) = uk : k x2 + · · · + xn k k . x2 + · · · + xn ≥ (n − 1) n−1 Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page Case 0 < x1 ≤ x2 = x3 = · · · = xn . Denoting x1 = x and x2 = x3 = · · · = xn = y, we have to prove that for 0 < x ≤ 1 ≤ y and x + (n − 1)y = n, the inequality holds: xk + (n − 1)y k + nrxy n−1 − n(r + 1) ≥ 0. JJ II Write the inequality as f (x) ≥ 0, where J I f (x) = xk + (n − 1)y k + nrxy n−1 − n(r + 1), We see that f (0) = f (1) = 0. Since y 0 = −1 , n−1 with y= n−x . n−1 = (y − x)y n−2 [nr − kg(x)], where y n−k + x y n−k+1 Go Back Full Screen = (y − x)[nry n−2 − k(y k−2 + y k−3 x + · · · + xk−2 )] 1 Page 24 of 41 we have f 0 (x) = k(xk−1 − y k−1 ) + nry n−2 (y − x) g(x) = Contents + ··· + xk−2 . y n−2 Close Since the function y(x) = n−x is strictly decreasing, the function g(x) is strictly n−1 increasing for 2 ≤ k ≤ n. For k = n + 1, we have xn−1 x2 + · · · + n−2 y y 2 (n − 2)x + n x xn−1 = + + · · · + n−2 , n−1 y y g(x) = y + x + and for k = n + 2, we have Equal Variable Method Vasile Cîrtoaje x3 xn + · · · + n−2 y y 2 2 (n − 3n + 3)x + n(n − 3)x + n2 x3 xn = + + · · · + . (n − 1)2 y y n−2 vol. 8, iss. 1, art. 15, 2007 g(x) = y 2 + yx + x2 + Title Page Contents Therefore, the function g(x) is strictly increasing for 2 ≤ k ≤ n+2, and the function JJ II h(x) = nr − kg(x) J I is strictly decreasing. Note that f 0 (x) = (y − x)y n−2 h(x). Page 25 of 41 Go Back Full Screen We assert that h(0) > 0 and h(1) < 0. If our claim is true, then there exists x1 ∈ (0, 1) such that h(x1 ) = 0, h(x) > 0 for x ∈ [0, x1 ), and h(x) < 0 for x ∈ (x1 , 1]. Consequently, f (x) is strictly increasing for x ∈ [0, x1 ], and strictly decreasing for x ∈ [x1 , 1]. Since f (0) = f (1) = 0, it follows that f (x) ≥ 0 for 0 < x ≤ 1, and the proof is completed. In order to prove that h(0) > 0, we assume that h(0) ≤ 0. Then, h(x) < 0 for x ∈ (0, 1), f 0 (x) < 0 for x ∈ (0, 1), and f (x) is strictly decreasing for x ∈ [0, 1], Close which contradicts f (0) = f (1). Also, if h(1) ≥ 0, then h(x) > 0 for x ∈ (0, 1), f 0 (x) > 0 for x ∈ (0, 1), and f (x) is strictly increasing for x ∈ [0, 1], which also contradicts f (0) = f (1). For n ≥ 3 and x1 ≤ x2 ≤ · · · ≤ xn , equality occurs when x1 = x2 = · · · = xn = n 1, and also when x1 = 0 and x2 = · · · = xn = n−1 . Remark 2. For k = 2, k = 3 and k = 4, we get the following nice inequalities: Equal Variable Method (n − 1)(x21 + x22 + · · · + x2n ) + nx1 x2 · · · xn ≥ n2 , Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 (n − 1)2 (x31 + x32 + · · · + x3n ) + n(2n − 1)x1 x2 · · · xn ≥ n3 , (n − 1)3 (x41 + x42 + · · · + x4n ) + n(3n2 − 3n + 1)x1 x2 · · · xn ≥ n4 . Title Page Remark 3. The inequality for k = n was posted in 2004 on the Mathlinks Site Inequalities Forum by Gabriel Dospinescu and Călin Popa. Proposition 3.6 ([11]). Let x1 , x2 , . . . , xn be positive real numbers such that 1 + · · · + x1n = n. Then x2 where en−1 1 x1 + x1 + x2 + · · · + xn − n ≤ en−1 (x1 x2 · · · xn − 1), n−1 1 = 1 + n−1 < e. Proof. Replacing each of the xi by a1i , the statement becomes as follows: If a1 , a2 , . . . , an are positive numbers such that a1 + a2 + · · · + an = n, then 1 1 1 a1 a2 · · · an + + ··· + − n + en−1 ≤ en−1 . a1 a2 an It is easy to check that the inequality holds for n = 2. Consider now n ≥ 3, assume that 0 < a1 ≤ a2 ≤ · · · ≤ an and apply Corollary 1.7 (case p = −1): If 0 < a1 ≤ Contents JJ II J I Page 26 of 41 Go Back Full Screen Close a2 ≤ · · · ≤ an such that a1 + a2 + · · · + an = n and a11 + a12 + · · · + a1n = constant, then the product a1 a2 · · · an is maximal when 0 < a1 ≤ a2 = a3 = · · · = an . Denoting a1 = x and a2 = a3 = · · · = an = y, we have to prove that for n 0 < x ≤ 1 ≤ y < n−1 and x + (n − 1)y = n, the inequality holds: y n−1 + (n − 1)xy n−2 − (n − en−1 )xy n−1 ≤ en−1 . Letting Equal Variable Method n−1 f (x) = y + (n − 1)xy n−x y= , n−1 n−2 − (n − en−1 )xy n−1 − en−1 , with we must show that f (x) ≤ 0 for 0 < x ≤ 1. We see that f (0) = f (1) = 0. Since −1 y 0 = n−1 , we have 0 f (x) = (y − x)[n − 2 − (n − en−1 )y] = (y − x)h(x), y n−3 where h(x) = n − 2 − (n − en−1 ) Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 n−x n−1 is a linear increasing function. Let us show that h(0) < 0 and h(1) > 0. If h(0) ≥ 0, then h(x) > 0 for x ∈ (0, 1), hence f 0 (x) > 0 for x ∈ (0, 1), and f (x) is strictly increasing for x ∈ [0, 1], which contradicts f (0) = f (1). Also, h(1) = en−1 − 2 > 0. From h(0) < 0 and h(1) > 0, it follows that there exists x1 ∈ (0, 1) such that h(x1 ) = 0, h(x) < 0 for x ∈ [0, x1 ), and h(x) > 0 for x ∈ (x1 , 1]. Consequently, f (x) is strictly decreasing for x ∈ [0, x1 ], and strictly increasing for x ∈ [x1 , 1]. Since f (0) = f (1) = 0, it follows that f (x) ≤ 0 for 0 ≤ x ≤ 1. For n ≥ 3, equality occurs when x1 = x2 = · · · = xn = 1. Title Page Contents JJ II J I Page 27 of 41 Go Back Full Screen Close Proposition 3.7 ([9]). If x1 , x2 , . . . , xn are positive real numbers, then xn1 + xn2 + · · · + xnn + n(n − 1)x1 x2 · · · xn ≥ x1 x2 · · · xn (x1 + x2 + · · · + xn ) 1 1 1 + + ··· + x1 x2 xn . Proof. For n = 2, one has equality. Assume now that n ≥ 3, 0 < x1 ≤ x2 ≤ · · · ≤ xn and apply Corollary 1.8 (case p = 0): If 0 < x1 ≤ x2 ≤ · · · ≤ xn such that x1 + x2 + · · · + xn = constant and x1 x2 · · · xn = constant, then the sum xn1 +xn2 +· · ·+xnn is minimal and the sum x11 + x12 +· · ·+ x1n is maximal when 0 < x1 ≤ x2 = x3 = · · · = xn . Thus, it suffices to prove the inequality for 0 < x1 ≤ 1 and x2 = x3 = · · · = xn = 1. The inequality becomes xn1 + (n − 2)x1 ≥ (n − 1)x21 , Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page Contents JJ II J I Page 28 of 41 and is equivalent to Go Back x1 (x1 − 1)[(xn−2 1 − 1) + (xn−3 1 − 1) + · · · + (x1 − 1)] ≥ 0, which is clearly true. For n ≥ 3, equality occurs if and only if x1 = x2 = · · · = xn . Proposition 3.8 ([14]). If x1 , x2 , . . . , xn are non-negative real numbers, then (n − 1)(xn1 + xn2 + · · · + xnn ) + nx1 x2 · · · xn + xn−1 + · · · + xn−1 ≥ (x1 + x2 + · · · + xn )(xn−1 1 2 n ). Full Screen Close Proof. For n = 2, one has equality. For n ≥ 3, assume that 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn and apply Corollary 1.8 (case p = n and q = n − 1) and Corollary 1.7 (case p = n): If 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn such that x1 + x2 + · · · + xn = constant and xn1 + xn2 + · · · + xnn = constant, then the sum xn−1 + xn−1 + · · · + xn−1 is maximal and the product x1 x2 · · · xn is 1 2 n minimal when either x1 = 0 or 0 < x1 ≤ x2 = x3 = · · · = xn . So, it suffices to consider the cases x1 = 0 and 0 < x1 ≤ x2 = x3 = · · · = xn . Case x1 = 0. The inequality reduces to Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 (n − 1)(xn2 + · · · + xnn ) ≥ (x2 + · · · + xn )(xn−1 + · · · + xn−1 2 n ), Title Page which immediately follows by Chebyshev’s inequality. Case 0 < x1 ≤ x2 = x3 = · · · = xn . Setting x2 = x3 = · · · = xn = 1, the inequality reduces to: Contents (n − 2)xn1 + x1 ≥ (n − 1)xn−1 . 1 x1 (x1 − II J I Page 29 of 41 Rewriting this inequality as 1)[xn−3 (x1 1 JJ Go Back − 1) + xn−4 (x21 1 − 1) + · · · + (xn−2 1 − 1)] ≥ 0, we see that it is clearly true. For n ≥ 3 and x1 ≤ x2 ≤ · · · ≤ xn equality occurs when x1 = x2 = · · · = xn , and for x1 = 0 and x2 = · · · = xn . Proposition 3.9 ([8]). If x1 , x2 , . . . , xn are positive real numbers, then 1 1 1 1 (x1 +x2 +· · ·+xn −n) + + ··· + − n +x1 x2 · · · xn + ≥ 2. x1 x2 xn x1 x2 · · · xn Full Screen Close Proof. For n = 2, the inequality reduces to (1 − x1 )2 (1 − x2 )2 ≥ 0. x1 x2 For n ≥ 3, assume that 0 < x1 ≤ x2 ≤ · · · ≤ xn . Since the inequality preserves its form by replacing each number xi with x1i , we may consider x1 x2 · · · xn ≥ 1. So, by the AM-GM inequality we get √ x1 + x2 + · · · + xn − n ≥ n n x1 x2 · · · xn − n ≥ 0, and we may apply Corollary 1.8 (case p = 0 and q = −1): If 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn such that x1 + x2 + · · · + xn = constant and x1 x2 · · · xn = constant, then the sum x11 + x12 + · · · + x1n is minimal when 0 < x1 = x2 = · · · = xn−1 ≤ xn . According to this statement, it suffices to consider x1 = x2 = · · · = xn−1 = x and xn = y, when the inequality reduces to n−1 1 1 ((n − 1)x + y − n) + − n + xn−1 y + n−1 ≥ 2, x y x y or n−1 1 1 n(n − 1)(x − 1)2 n−1 x + − n y + n−1 + (n − 1)x − n ≥ . x x y x Since n−1 x − 1 n−1 −n= [(x − 1) + (xn−2 − 1) + · · · + (x − 1)] xn−1 + x x (x − 1)2 n−2 = [x + 2xn−3 + · · · + (n − 1)] x Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page Contents JJ II J I Page 30 of 41 Go Back Full Screen Close and 1 xn−1 (x − 1)2 1 2 + (n − 1)x − n = + + · · · + (n − 1) , x xn−2 xn−3 it is enough to show that [x n−2 + 2x n−3 + · · · + (n − 1)]y + 1 xn−2 + 2 xn−3 1 + · · · + (n − 1) ≥ n(n − 1). y This inequality is equivalent to 1 1 n−2 n−3 x y + n−2 − 2 + 2 x y + n−3 − 2 x y x y 1 + · · · + (n − 1) y + − 2 ≥ 0, y or (xn−2 y − 1)2 2(xn−3 y − 1)2 (n − 1)(y − 1)2 + + · · · + ≥ 0, xn−2 y xn−3 y y which is clearly true. Equality occurs if and only if n − 1 of the numbers xi are equal to 1. Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page Contents JJ II J I Page 31 of 41 Go Back Full Screen Proposition 3.10 ([15]). If x1 , x2 , . . . , xn are non-negative real numbers such that x1 + x2 + · · · + xn = n, then (x1 x2 · · · xn ) √1 n−1 (x21 + x22 + · · · + x2n ) ≤ n. Proof. For n = 2, the inequality reduces to 2(x1 x2 − 1)2 ≥ 0. For n ≥ 3, assume that 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn and apply Corollary 1.7 (case p = 2): If 0 ≤ x1 ≤ Close x2 ≤ · · · ≤ xn such that x1 + x2 + · · · + xn = n and x21 + x22 + · · · + x2n = constant, then the product x1 x2 · · · xn is maximal when 0 ≤ x1 = x2 = · · · = xn−1 ≤ xn . Consequently, it suffices to show that the inequality holds for x1 = x2 = · · · = xn−1 = x and xn = y, where 0 ≤ x ≤ 1 ≤ y and (n − 1)x + y = n. Under the circumstances, the inequality reduces to √ x n−1 y √1 n−1 [(n − 1)x2 + y 2 ] ≤ n. Equal Variable Method For x = 0, the inequality is trivial. For x > 0, it is equivalent to f (x) ≤ 0, where f (x) = √ Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 n − 1 ln x + √ 1 ln y + ln[(n − 1)x2 + y 2 ] − ln n, n−1 with y = n − (n − 1)x. Title Page Contents We have y 0 = −(n − 1) and √ √ 1 1 2 n − 1(x − y) f 0 (x) (y − x)( n − 1x − y)2 √ = − + = ≥ 0. x y (n − 1)x2 + y 2 xy[(n − 1)x2 + y 2 ] n−1 JJ II J I Therefore, the function f (x) is strictly increasing on (0, 1] and hence f (x) ≤ f (1) = 0. Equality occurs if and only if x1 = x2 = · · · = xn = 1. Page 32 of 41 Remark 4. For n = 5, we get the following nice statement: If a, b, c, d, e are positive real numbers such that a2 + b2 + c2 + d2 + e2 = 5, then Full Screen 4 4 4 4 4 abcde(a + b + c + d + e ) ≤ 5. Proposition 3.11 ([4]). Let x, y, z be non-negative real numbers such that xy + yz + zx = 3, and let ln 9 − ln 4 p≥ ≈ 0.738. ln 3 Go Back Close Then, xp + y p + z p ≥ 3. Proof. Let r = ln 9−ln 4 . ln 3 By the Power-Mean inequality, we have xp + y p + z p ≥ 3 xr + y r + z r 3 pr . Equal Variable Method Thus, it suffices to show that Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 xr + y r + z r ≥ 3. Let x ≤ y ≤ z. We consider two cases. Case x = 0. We have to show that y r + z r ≥ 3 for yz = 3. Indeed, by the AM-GM inequality, we get y r + z r ≥ 2(yz)r/2 = 2 · 3r/2 = 3. r r r Case x > 0. The inequality x + y + z ≥ 3 is equivalent to the homogeneous inequality xyz r2 1 1 1 r2 r r r x +y +z ≥3 + + . 3 x y z 1 r 1 r 1 r Setting x = a , y = b , z = c (0 < a ≤ b ≤ c), the inequality becomes a+b+c≥3 abc 3 12 a −1 r +b −1 r +c −1 r r2 Title Page Contents JJ II J I Page 33 of 41 Go Back Full Screen Close . Towards proving this inequality, we apply Corollary 1.8 (case p = 0, q = −1 ): If r 0 < a ≤ b ≤ c such that a + b + c = constant and abc = constant, then the sum −1 −1 −1 a r + b r + c r is maximal when 0 < a ≤ b = c. So, it suffices to prove the inequality for 0 < a ≤ b = c; that is, to prove the homogeneous inequality in x, y, z for 0 < x ≤ y = z. Without loss of generality, we may leave aside the constraint xy + yz + zx = 3, and consider y = z = 1 and 0 < x ≤ 1. The inequality reduces to r 2x + 1 2 r x +2≥3 . 3 Equal Variable Method Denoting xr + 2 r 2x + 1 f (x) = ln − ln , 3 2 3 we have to show that f (x) ≥ 0 for 0 < x ≤ 1. The derivative rxr−1 r r(x − 2x1−r + 1) f 0 (x) = r − = 1−r r x + 2 2x + 1 x (x + 2)(2x + 1) has the same sign as g(x) = x − 2x1−r + 1. Since g 0 (x) = 1 − 2(1−r) , we see that xr 0 0 g (x) < 0 for x ∈ (0, x1 ), and g (x) > 0 for x ∈ (x1 , 1], where x1 = (2 − 2r)1/r ≈ 0.416. The function g(x) is strictly decreasing on [0, x1 ], and strictly increasing on [x1 , 1]. Since g(0) = 1 and g(1) = 0, there exists x2 ∈ (0, 1) such that g(x2 ) = 0, g(x) > 0 for x ∈ [0, x2 ) and g(x) < 0 for x ∈ (x2 , 1). Consequently, the function f (x) is strictly increasing on [0, x2 ] and strictly decreasing on [x2 , 1]. Since f (0) = f (1) = 0, we have f (x) ≥ 0 for 0 < x ≤ 1, establishing the desired result. 4 Equality occurs for x = y = z = 1. Additionally, for p = ln 9−ln and x ≤ y ≤ z, ln 3 √ equality holds again for x = 0 and y = z = 3. Proposition 3.12 ([7]). Let x, y, z be non-negative real numbers such that x+y+z = ln 9−ln 8 3, and let p ≥ ln ≈ 0.29. Then, 3−ln 2 xp + y p + z p ≥ xy + yz + zx. Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page Contents JJ II J I Page 34 of 41 Go Back Full Screen Close Proof. For p ≥ 1, by Jensen’s inequality we have p x+y+z p p p x +y +z ≥3 3 1 = 3 = (x + y + z)2 ≥ xy + yz + zx. 3 9−ln 8 Assume now p < 1. Let r = ln and x ≤ y ≤ z. The inequality is equivalent to ln 3−ln 2 the homogeneous inequality 2−p x+y+z p p p + x2 + y 2 + z 2 ≥ (x + y + z)2 . 2(x + y + z ) 3 By Corollary 1.8 (case 0 < p < 1 and q = 2), if x ≤ y ≤ z such that x + y + z = constant and xp + y p + z p = constant, then the sum x2 + y 2 + z 2 is minimal when either x = 0 or 0 < x ≤ y = z. Case x = 0. Returning to our original inequality, we have to show that y p + z p ≥ yz for y + z = 3. Indeed, by the AM-GM inequality, we get p p 2 p y + z − yz ≥ 2(yz) − yz p Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page Contents JJ II J I Page 35 of 41 2−p = (yz) 2 [2 − (yz) 2 ] " 2−p # p y+z ≥ (yz) 2 2 − 2 " 2−p # p 3 = (yz) 2 2 − 2 " 2−r # p 3 = 0. ≥ (yz) 2 2 − 2 Go Back Full Screen Close Case 0 < x ≤ y = z. In the homogeneous inequality, we may leave aside the constraint x + y + z = 3, and consider y = z = 1 and 0 < x ≤ 1. Thus, the inequality reduces to 2−p x+2 p (x + 2) ≥ 2x + 1. 3 To prove this inequality, we consider the function x+2 f (x) = ln(x + 2) + (2 − p) ln − ln(2x + 1). 3 p We have to show that f (x) ≥ 0 for 0 < x ≤ 1 and r ≤ p < 1. We have f 0 (x) = Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page p−1 2−p 2 2g(x) px + − = 1−p p , p x +2 x + 2 2x + 1 x (x + 2)(2x + 1) where g(x) = x2 + (2p − 1)x + p + 2(1 − p)x2−p − (p + 2)x1−p , Contents JJ II J I Page 36 of 41 and g 0 (x) = 2x + 2p − 1 + 2(1 − p)(2 − p)x1−p − (p + 2)(1 − p)x−p , Go Back g 00 (x) = 2 + 2(1 − p)2 (2 − p)x−p + p(p + 2)(1 − p)x−p−1 . Full Screen Since g 00 (x) > 0, the first derivative g 0 (x) is strictly increasing on (0, 1]. Taking into account that g 0 (0+) = −∞ and g 0 (1) = 3(1 − p) + 3p2 > 0, there is x1 ∈ (0, 1) such that g 0 (x1 ) = 0, g 0 (x) < 0 for x ∈ (0, x1 )and g 0 (x) > 0 for x ∈ (x1 , 1]. Therefore, the function g(x) is strictly decreasing on [0, x1 ] and strictly increasing on [x1 , 1]. Since g(0) = p > 0 and g(1) = 0, there is x2 ∈ (0, x1 ) such that g(x2 ) = 0, g(x) > 0 for x ∈ [0, x2 ) and g(x) < 0 for x ∈ (x2 , 1]. We have also f 0 (x2 ) = 0, Close f 0 (x) > 0 for x ∈ (0, x2 ) and f 0 (x) < 0 for x ∈ (x2 , 1]. According to this result, the function f (x) is strictly increasing on [0, x2 ] and strictly decreasing on [x2 , 1]. Since f (0) = ln 2 + (2 − p) ln 2 2 ≥ ln 2 + (2 − r) ln = 0 3 3 and f (1) = 0, we get f (x) ≥ min{f (0), f (1)} = 0. Equality occurs for x = y = z = 1. Additionally, for p = equality holds again when x = 0 and y = z = 32 . ln 9−ln 8 ln 3−ln 2 and x ≤ y ≤ z, Equal Variable Method Vasile Cîrtoaje Proposition 3.13 ([8]). If x1 , x2 , . . . , xn (n ≥ 4) are non-negative numbers such that x1 + x2 + · · · + xn = n, then vol. 8, iss. 1, art. 15, 2007 1 1 1 + + ··· + ≤ 1. n + 1 − x2 x3 · · · xn n + 1 − x3 x4 · · · x1 n + 1 − x1 x2 · · · xn−1 n−1 1 Proof. Let x1 ≤ x2 ≤ · · · ≤ xn and en−1 = 1 + n−1 . By the AM-GM inequality, we have n−1 n−1 x2 + · · · + xn x1 + x2 + · · · + xn x2 · · · xn ≤ ≤ = en−1 . n−1 n−1 Title Page Hence Contents JJ II J I Page 37 of 41 Go Back n + 1 − x2 x3 · · · xn ≥ n + 1 − en−1 > 0, Full Screen and all denominators of the inequality are positive. Case x1 = 0. It is easy to show that the inequality holds. Case x1 > 0. Suppose that x1 x2 · · · xn = (n + 1)r = constant, r > 0. The inequality becomes x1 x2 xn + + ··· + ≤ n + 1, x1 − r x2 − r xn − r Close or 1 1 1 1 + + ··· + ≤ . x1 − r x2 − r xn − r r By the AM-GM inequality, we have n x1 + x2 + · · · + xn (n + 1)r = x1 x2 · · · xn ≤ = 1, n Equal Variable Method 1 . n+1 1 , r 1 . r hence r ≤ From xn < x1 + x2 + · · · + xn = n < n + 1 ≤ we get xn < Therefore, we have r < xi < 1r for all numbers xi . −1 We will apply now Corollary 1.6 to the function f (u) = u−r , u > r. We have 1 0 f (u) = (u−r)2 and 1 4rx + 2 x2 0 g(x) = f = , g 00 (x) = . 2 x (1 − rx) (1 − rx)4 Since g 00 (x) > 0, g(x) is strictly convex on r, 1r . According to Corollary 1.6, if 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn such that for x1 + x2 + · · · + xn = constant and x1 x2 · · · xn = constant, then the sum f (x1 ) + f (x2 ) + · · · + f (xn ) is minimal when x1 ≤ x2 = x3 = · · · = xn . Thus, to prove the original inequality, it suffices to consider the case x1 = x and x2 = x3 = · · · = xn = y, where 0 < x ≤ 1 ≤ y and x + (n − 1)y = n. We leave ending the proof to the reader. Remark 5. The inequality is a particular case of the following more general statement: n−1 1 n−1 Let n ≥ 3, en−1 = 1 + n−1 , kn = (n−1)e and let a1 , a2 , . . . , an be nonn−en−1 negative numbers such that a1 + a2 + · · · + an = n. Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page Contents JJ II J I Page 38 of 41 Go Back Full Screen Close (a) If k ≥ kn , then 1 1 1 n + + ··· + ≤ ; k − a2 a2 · · · an k − a3 a4 · · · a1 k − a1 a2 · · · an−1 k−1 (b) If en−1 < k < kn , then 1 1 n−1 1 1 + +· · ·+ ≤ + . k − a2 a3 · · · an k − a3 a4 · · · a1 k − a1 a2 · · · an−1 k k − en−1 Finally, we mention that many other applications of the EV-Method are given in the book [2]. Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page Contents JJ II J I Page 39 of 41 Go Back Full Screen Close References [1] V. CÎRTOAJE, A generalization of Jensen’s inequality, Gazeta Matematică A, 2 (2005), 124–138. [2] V. CÎRTOAJE, Algebraic Inequalities - Old and New Methods, GIL Publishing House, Romania, 2006. [3] V. CÎRTOAJE, Two generalizations of Popoviciu’s Inequality, Crux Math., 31(5) (2005), 313–318. [4] V. CÎRTOAJE, Solution to problem 2724 by Walther Janous, Crux Math., 30(1) (2004), 44–46. Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page [5] V. CÎRTOAJE, Problem 10528, A.M.M, 103(5) (1996), 428. [6] V. CÎRTOAJE, Mathlinks site, http://www.mathlinks.ro/Forum/ viewtopic.php?t=123604, Dec 10, 2006. [7] V. CÎRTOAJE, Mathlinks site, http://www.mathlinks.ro/Forum/ viewtopic.php?t=56732, Oct 19, 2005. [8] V. CÎRTOAJE, Mathlinks site, http://www.mathlinks.ro/Forum/ viewtopic.php?t=18627, Jan 29, 2005. Contents JJ II J I Page 40 of 41 Go Back Full Screen [9] V. CÎRTOAJE, Mathlinks site, http://www.mathlinks.ro/Forum/ viewtopic.php?t=14906, Aug 04, 2004. [10] V. CÎRTOAJE, Mathlinks site, http://www.mathlinks.ro/Forum/ viewtopic.php?t=19076, Nov 2, 2004. [11] V. CÎRTOAJE, Mathlinks site, http://www.mathlinks.ro/Forum/ viewtopic.php?t=21613, Feb 15, 2005. Close [12] W. JANOUS AND V. CÎRTOAJE, Two solutions to problem 2664, Crux Math., 29(5) (2003), 321–323. [13] D.S. MITRINOVIC, J.E. PECARIC equalities in Analysis, Kluwer, 1993. AND A.M. FINK, Classical and New In- [14] J. SURANYI, Miklos Schweitzer Competition-Hungary. http://www. mathlinks.ro/Forum/viewtopic.php?t=14008, Jul 11, 2004. [15] http://www.mathlinks.ro/Forum/viewtopic.php?t=36346, May 10, 2005. [16] http://www.mathlinks.ro/Forum/viewtopic.php?p= 360537, Nov 2, 2005. Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007 Title Page Contents JJ II J I Page 41 of 41 Go Back Full Screen Close