Volume 8 (2007), Issue 1, Article 15, 21 pp.
THE EQUAL VARIABLE METHOD
VASILE CÎRTOAJE
D EPARTMENT OF AUTOMATION AND C OMPUTERS
U NIVERSITY OF P LOIE ŞTI
B UCURE ŞTI 39, ROMANIA
vcirtoaje@upg-ploiesti.ro
Received 01 March, 2006; accepted 17 April, 2006
Communicated by P.S. Bullen
A BSTRACT. 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 ).
Key words and phrases: Symmetric inequalities, Power means, EV-Theorem.
2000 Mathematics Subject Classification. 26D10, 26D20.
1. S TATEMENT 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,
(1)
(2)
(3)
(4)
if x = x1 and p ≤ 0, then 0 < x < y = z;
if x = x1 and p > 1, then either 0 = x < y ≤ z or 0 < x < y = z;
if x ∈ (x1 , x2 ), then x < y < z;
if x = x2 , then x = y < z.
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,
059-06
x p + y p + z p = ap + b p + c p ,
2
VASILE C ÎRTOAJE
where p ∈ (−∞, 0] ∪ (1, ∞). For p = 0, the second equation
1 is xyz = abc > 0. Let f (u) be a
0
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 is maximal
u→0
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
x 1 + x 2 + · · · + x n = a1 + a2 + · · · + an ,
xp1 + xp2 + · · · + xpn = ap1 + ap2 + · · · + apn ,
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
is strictly convex on (0, ∞), and let
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 < x 1 ≤ x2 = x3 = · · · = xn ;
(2) If p > 0 and either f (u) is continuous at u = 0 or lim f (u) = −∞, then Fn is
u→0
maximal for 0 ≤ x1 = x2 = · · · = xn−1 ≤ xn , and is minimal for either x1 = 0 or
0 < x 1 ≤ x2 = x3 = · · · = xn .
Remark 1.4.
0 < α < β. If the function f is differentiable on (α, β) and the function
Let
1
0
g(x) = f 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).
By Theorem 1.3, we easily obtain some particular results, which are very useful in applications.
Corollary 1.5. 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 .
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,
x→0
Fn = f (x1 ) + f (x2 ) + · · · + f (xn )
is maximal for 0 ≤ x1 = x2 = · · · = xn−1 ≤ xn , and is minimal for either x1 = 0 or
0 < x 1 ≤ x2 = x3 = · · · = xn .
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 15, 21 pp.
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E QUAL VARIABLE M ETHOD
3
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 ,
1
1
1
1
1
1
+
+ ··· +
=
+
+ ··· + .
x1 x2
xn
a1 a2
an
0 √1
Let f be a differentiable function on (0, ∞) such that g(x) = f
is strictly convex on
x
(0, ∞). Then,
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 .
Corollary 1.7. 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 )
is maximal for 0 < x1 = x2 = · · · = xn−1 ≤ xn , and is minimal for 0 < x1 ≤ x2 = x3 = · · · =
xn .
Corollary 1.8. Let a1 , a2 , . . . , an (n ≥ 3) be fixed non-negative numbers, and let 0 ≤ x1 ≤
x2 ≤ · · · ≤ xn such that
x1 + x2 + · · · + xn = 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 1.9. 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 .
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.
(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 .
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 15, 21 pp.
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4
VASILE C ÎRTOAJE
(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 .
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 .
2. P ROOFS
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
x p + y p + z p = ap + b p + c p .
Consider now three cases: p = 0, p < 0 and p > 1.
√
3
A. Case p = 0 (xyz = abc > 0). Let S = a+b+c
and
P
=
abc, where S > P > 0 by AM-GM
3
Inequality. We have
x + y + z = 3S, xyz = P 3 ,
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.
Sub-case x > x1 . We have f (x) > 0 and y < z. Consider now that y and z depend on x. From
0
0
x + y(x) + z(x) = 3S and x · y(x) · z(x) = P 3 , we get 1 + y 0 + z 0 = 0 and x1 + yy + zz = 0.
Hence,
y(x − z)
z(y − x)
y 0 (x) =
, z 0 (x) =
.
x(z − y)
x(z − y)
Since y 0 (x) < 0, the function y(x) is strictly decreasing. Since 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 . 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 .
p
p
p 1
B. Case p < 0. Denote S = a+b+c
and R = a +b3 +c p . Taking into account that
3
x + y + z = 3S,
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 15, 21 pp.
xp + y p + z p = 3Rp ,
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E QUAL VARIABLE M ETHOD
5
1
from 0 < x ≤ y ≤ z and x < z we get x < S and 3 p R < x < R. Let
−1
p
y + zp p
− 2.
h = (y + z)
2
By the AM-GM Inequality, we have
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
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 equation
3Rp
h0 (x) =
2
3Rp − xp
2
1
h(x) = 0 has a unique root x1 and x ≥ x1 > 3 p R.
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 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 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, and hence
x < S < R. Let
p
1
2
y + zp p
h=
− 1.
y+z
2
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
and
p
1−p
3
3R − xp p
h (x) =
(Rp − Sxp−1 ) > 0,
(3S − x)2
2
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.
0
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 15, 21 pp.
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6
VASILE C ÎRTOAJE
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) =
xp−1 − z p−1
,
z p−1 − y p−1
z 0 (x) =
y p−1 − xp−1
.
z p−1 − 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 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
p
p
p
p
p
and
p
x + y (x) + z (x) = a + b + c ,
we get
y 0 + z 0 = −1,
y p−1 y 0 + z p−1 z 0 = −xp−1 ,
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)
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.
These results imply F 0 (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
u→0
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E QUAL VARIABLE M ETHOD
7
F (x) is continuous on [x1 , x2 ], and hence is minimal 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) = −∞,
u→0
the function Fn (x1 , x2 , . . . , xn ) attains its minimum and maximum 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 )
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 , each of
1 1
the xi by xip , and then p by p1 . Thus, we obtain the sufficient condition that h(x) = xf 0 x 1−p
to be strictly
convex
on (0, ∞). We claim that this condition is equivalent to the condition that
1
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,
1
1
we see that g x = x 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, ∞).
Proof of Corollary 1.8. We will apply Theorem 1.3 to the function f (u) = p ln u. We see that
lim f (u) = −∞ for p > 0, and
u→0
p
f (u) = ,
u
0
g(x) = f
0
x
1
p−1
= px
1
1−p
,
2p−1
p2
1−p .
g (x) =
x
(1 − p)2
00
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 1.9. 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
f 0 (u) = q 2 (q − 1)(q − p)uq−1
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8
VASILE C ÎRTOAJE
and
1 q−1
g(x) = f 0 x p−1 = q 2 (q − 1)(q − p)x p−1 ,
q 2 (q − 1)2 (q − p)2 2p−1
g (x) =
x 1−p .
(p − 1)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.
00
3. A PPLICATIONS
Proposition 3.1. Let x, y, z be non-negative real numbers such that x+y+z = 2. If r0 ≤ r ≤ 3,
ln 2
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 1.9 (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 + z2 2
y r−1 + z r−1
≤
.
2
2
Thus, it suffices to show that
2
r−1
y + z2 2
yz
≤ 1.
2
Taking account of
y2 + z2
2(y 2 + z 2 )
r−1
=
≥ 1 and
≤ 1,
2
2
(y + z)
2
we have
2
2
r−1
y + z2 2
y + z2
1 − yz
≥ 1 − yz
2
2
4
(y + z)
yz(y 2 + z 2 )
=
−
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
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E QUAL VARIABLE M ETHOD
Since 1 +
Let
x r+1
2
9
is increasing and xr is decreasing in respect to r, it suffices to consider r = r0 .
x r0 +1
− xr0 − x − 1.
f (x) = 1 +
2
We have
x r0
r0 + 1 1+
− r0 xr0 −1 − 1,
2
2
1 00
r0 + 1 x r0 r0 − 1
f (x) =
1+
− 2−r0 .
r0
4
2
x
f 0 (x) =
Since f 00 (x) is strictly increasing on (0, 1], f 00 (0+ ) = −∞ and
r
r0 + 1 3 0
1 00
f (1) =
− r0 + 1
r0
4
2
3 − r0
r0 + 1
=
− 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
r0 + 1
3
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.
Proposition 3.2 ([12]). Let x, y, z be non-negative real numbers such that xy + yz + zx = 3.
If 1 < r ≤ 2, then
xr (y + z) + y r (z + x) + z r (x + y) ≥ 6.
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 1.5 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 1.5,
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.
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 15, 21 pp.
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10
VASILE C ÎRTOAJE
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.
2
Let f (x) = xr (x + z) + xz r − 3, with z = 3−x
.
2x
2
Differentiating the equation x + 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
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 1.6 to the function
−(n−1)
1
f (u) = 1+(n−1)u
for u > 0. We have f 0 (u) = [1+(n−1)u]
2 and
1
−(n − 1)x
0
g(x) = f √
= √
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 1.6, 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
The last constraint is equivalent to
(n − 1)(y − 1) =
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 15, 21 pp.
y(1 − x2 )
.
x(1 + y)
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E QUAL VARIABLE M ETHOD
11
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],
which reduces to
(y − x)[(n − 1)xy − 1] ≥ 0.
Since y − x ≥ 0, we have still to prove that
(n − 1)xy ≥ 1.
Indeed, from x + (n − 1)y =
1
x
+
n−1
y
we get xy =
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.
(n − 1)xy − 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
m
am
1 + a2 + 2m − 2 ≥ m(a1 + a2 ).
m
We can prove it by summing the inequalities am
1 ≥ 1+m(a1 −1) and a2 ≥ 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
1
1
1
+
+
·
·
·
+
+ (m − 1)n ≥ m(x1 + x2 + · · · + xn )
xm
xm
xm
1
2
n
for x1 x2 · · · xn = 1. Assume 0 < x1 ≤ x2 ≤ · · · ≤ xn and apply Corollary 1.9 (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
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 15, 21 pp.
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12
VASILE C ÎRTOAJE
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
This inequality is equivalent to
xmn−m − 1 − m(n − 1)(x − 1) ≥ 0
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 +· · ·+
k−1
n
xn = n. If k is a positive integer satisfying 2 ≤ k ≤ n + 2, and r = n−1
− 1, then
xk1 + xk2 + · · · + xkn − n ≥ nr(1 − x1 x2 · · · xn ).
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 1.8 (case p = k > 0): If 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn 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
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.
Write the inequality as f (x) ≥ 0, where
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
we have
f 0 (x) = k(xk−1 − y k−1 ) + nry n−2 (y − x)
= (y − x)[nry n−2 − k(y k−2 + y k−3 x + · · · + xk−2 )]
= (y − x)y n−2 [nr − kg(x)],
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 15, 21 pp.
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E QUAL VARIABLE M ETHOD
13
where
g(x) =
1
y n−k
+
x
y n−k+1
+ ··· +
xk−2
.
y n−2
Since the function y(x) = n−x
is strictly decreasing, the function g(x) is strictly increasing for
n−1
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
x3
xn
+ · · · + n−2
y
y
2
2
xn
(n − 3n + 3)x + n(n − 3)x + n2 x3
+
+ · · · + n−2 .
=
(n − 1)2
y
y
g(x) = y 2 + yx + x2 +
Therefore, the function g(x) is strictly increasing for 2 ≤ k ≤ n + 2, and the function
h(x) = nr − kg(x)
is strictly decreasing. Note that
f 0 (x) = (y − x)y n−2 h(x).
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),
0
f (x) < 0 for x ∈ (0, 1), and f (x) is strictly decreasing for x ∈ [0, 1], 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 = 1, and also
n
when x1 = 0 and x2 = · · · = xn = n−1
.
Remark 3.6. For k = 2, k = 3 and k = 4, we get the following nice inequalities:
(n − 1)(x21 + x22 + · · · + x2n ) + nx1 x2 · · · xn ≥ n2 ,
(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 .
Remark 3.7. The inequality for k = n was posted in 2004 on the Mathlinks Site - Inequalities
Forum by Gabriel Dospinescu and Călin Popa.
Proposition 3.8 ([11]). Let x1 , x2 , . . . , xn be positive real numbers such that x11 + x12 +· · ·+ x1n =
n. Then
x1 + x2 + · · · + xn − n ≤ en−1 (x1 x2 · · · xn − 1),
n−1
1
where en−1 = 1 + n−1
< e.
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 15, 21 pp.
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14
VASILE C ÎRTOAJE
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
+
+ ··· +
− n + en−1 ≤ en−1 .
a1 a2 · · · an
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 1.8 (case p = −1): If 0 < a1 ≤ 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 0 < x ≤ 1 ≤ y <
n
and x + (n − 1)y = n, the inequality holds:
n−1
y n−1 + (n − 1)xy n−2 − (n − en−1 )xy n−1 ≤ en−1 .
Letting
f (x) = y n−1 + (n − 1)xy n−2 − (n − en−1 )xy n−1 − en−1 , with
n−x
y=
,
n−1
−1
we must show that f (x) ≤ 0 for 0 < x ≤ 1. We see that f (0) = f (1) = 0. Since y 0 = n−1
, we
have
f 0 (x)
= (y − x)[n − 2 − (n − en−1 )y] = (y − x)h(x),
y n−3
where
n−x
h(x) = n − 2 − (n − en−1 )
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.
Proposition 3.9 ([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 1.9 (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 < x 1 ≤ 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 ,
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E QUAL VARIABLE M ETHOD
15
and is equivalent to
x1 (x1 − 1)[(xn−2
− 1) + (xn−3
− 1) + · · · + (x1 − 1)] ≥ 0,
1
1
which is clearly true. For n ≥ 3, equality occurs if and only if x1 = x2 = · · · = xn .
Proposition 3.10 ([14]). If x1 , x2 , . . . , xn are non-negative real numbers, then
(n − 1)(xn1 + xn2 + · · · + xnn ) + nx1 x2 · · · xn
≥ (x1 + x2 + · · · + xn )(xn−1
+ xn−1
+ · · · + xnn−1 ).
1
2
Proof. For n = 2, one has equality. For n ≥ 3, assume that 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn
and apply Corollary 1.9 (case p = n and q = n − 1) and Corollary 1.8 (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 minimal when
1
2
n
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
+ · · · + xnn−1 ),
(n − 1)(xn2 + · · · + xnn ) ≥ (x2 + · · · + xn )(xn−1
2
which immediately follows by Chebyshev’s inequality.
Case 0 < x1 ≤ x2 = x3 = · · · = xn . Setting x2 = x3 = · · · = xn = 1, the inequality reduces
to:
(n − 2)xn1 + x1 ≥ (n − 1)xn−1
.
1
Rewriting this inequality as
(x1 − 1) + xn−4
(x21 − 1) + · · · + (x1n−2 − 1)] ≥ 0,
x1 (x1 − 1)[xn−3
1
1
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 3.11 ([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
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 1.9 (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
1
x1
+
1
x2
+ ··· +
1
xn
is minimal when 0 < x1 = x2 = · · · = xn−1 ≤ xn .
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 15, 21 pp.
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16
VASILE C ÎRTOAJE
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
1
1
n−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
and
(x − 1)2
1
2
1
+ (n − 1)x − n =
+
+ · · · + (n − 1) ,
xn−1
x
xn−2 xn−3
it is enough to show that
1
2
1
n−2
n−3
[x
+ 2x
+ · · · + (n − 1)]y + n−2 + n−3 + · · · + (n − 1)
≥ n(n − 1).
x
x
y
This inequality is equivalent to
1
1
n−3
n−2
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.
Proposition 3.12 ([15]). If x1 , x2 , . . . , xn are non-negative real numbers such that x1 + x2 +
· · · + xn = n, then
√1
(x1 x2 · · · xn ) 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 1.8 (case p = 2): If 0 ≤ x1 ≤ 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
√
√1
x n−1 y n−1 [(n − 1)x2 + y 2 ] ≤ n.
For x = 0, the inequality is trivial. For x > 0, it is equivalent to f (x) ≤ 0, where
f (x) =
√
n − 1 ln x + √
1
ln y + ln[(n − 1)x2 + y 2 ] − ln n,
n−1
with y = n − (n − 1)x.
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E QUAL VARIABLE M ETHOD
17
We have y 0 = −(n − 1) and
√
√
f 0 (x)
1 1 2 n − 1(x − y)
(y − x)( n − 1x − y)2
√
= − +
=
≥ 0.
x y
(n − 1)x2 + y 2
xy[(n − 1)x2 + y 2 ]
n−1
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.
Remark 3.13. 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
abcde(a4 + b4 + c4 + d4 + e4 ) ≤ 5.
Proposition 3.14 ([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
Then,
xp + y p + z p ≥ 3.
Proof. Let r =
ln 9−ln 4
.
ln 3
By the Power-Mean inequality, we have
r
p
xp + y p + z p
x + yr + zr r
≥
.
3
3
Thus, it suffices to show that
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.
Case x > 0. The inequality xr + y r + z r ≥ 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
1
1
Setting x = a r , y = b r , z = c r (0 < a ≤ b ≤ c), the inequality becomes
1
r2
−1
−1
abc 2 −1
a+b+c≥3
a r +b r +c r
.
3
Towards proving this inequality, we apply Corollary 1.9 (case p = 0, q = −1
): If 0 < a ≤ b ≤ c
r
−1
−1
−1
such that a + b + c = constant and abc = constant, then the sum 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
r2
2x
+
1
xr + 2 ≥ 3
.
3
Denoting
f (x) = ln
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 15, 21 pp.
xr + 2 r 2x + 1
− ln
,
3
2
3
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18
VASILE C ÎRTOAJE
we have to show that f (x) ≥ 0 for 0 < x ≤ 1. The derivative
f 0 (x) =
rxr−1
r
r(x − 2x1−r + 1)
−
=
xr + 2 2x + 1
x1−r (xr + 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 g 0 (x) < 0
xr
for x ∈ (0, x1 ), and g 0 (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, equality
ln 3
√
holds again for x = 0 and y = z = 3.
Proposition 3.15 ([7]). Let x, y, z be non-negative real numbers such that x + y + z = 3, and
ln 9−ln 8
let p ≥ ln
≈ 0.29. Then,
3−ln 2
xp + y p + z p ≥ xy + yz + zx.
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 the
ln 3−ln 2
homogeneous inequality
2−p
x+y+z
p
p
p
2(x + y + z )
+ x2 + y 2 + z 2 ≥ (x + y + z)2 .
3
By Corollary 1.9 (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
y p + z p − yz ≥ 2(yz) 2 − yz
p
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
≥ (yz) 2 2 −
= 0.
2
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 15, 21 pp.
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E QUAL VARIABLE M ETHOD
19
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
− ln(2x + 1).
3
We have to show that f (x) ≥ 0 for 0 < x ≤ 1 and r ≤ p < 1. We have
f (x) = ln(xp + 2) + (2 − p) ln
f 0 (x) =
pxp−1
2−p
2
2g(x)
+
−
= 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 ,
and
g 0 (x) = 2x + 2p − 1 + 2(1 − p)(2 − p)x1−p − (p + 2)(1 − p)x−p ,
g 00 (x) = 2 + 2(1 − p)2 (2 − p)x−p + p(p + 2)(1 − p)x−p−1 .
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, 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
2
2
f (0) = ln 2 + (2 − p) ln ≥ ln 2 + (2 − r) ln = 0
3
3
and f (1) = 0, we get f (x) ≥ min{f (0), f (1)} = 0.
9−ln 8
Equality occurs for x = y = z = 1. Additionally, for p = ln
and x ≤ y ≤ z, equality
ln 3−ln 2
3
holds again when x = 0 and y = z = 2 .
Proposition 3.16 ([8]). If x1 , x2 , . . . , xn (n ≥ 4) are non-negative numbers such that x1 + x2 +
· · · + xn = n, then
1
1
1
+
+ ··· +
≤ 1.
n + 1 − x2 x3 · · · xn n + 1 − x3 x4 · · · x1
n + 1 − x1 x2 · · · xn−1
n−1
1
. By the AM-GM inequality, we have
Proof. Let x1 ≤ x2 ≤ · · · ≤ xn and en−1 = 1 + n−1
n−1 n−1
x2 + · · · + xn
x1 + x2 + · · · + xn
x2 · · · xn ≤
≤
= en−1 .
n−1
n−1
Hence
n + 1 − x2 x3 · · · xn ≥ n + 1 − en−1 > 0,
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
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 15, 21 pp.
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20
VASILE C ÎRTOAJE
or
1
1
1
1
+
+ ··· +
≤ .
x1 − r x2 − r
xn − r
r
By the AM-GM inequality, we have
n
x1 + x2 + · · · + xn
= 1,
(n + 1)r = x1 x2 · · · xn ≤
n
1
hence r ≤ n+1
. From xn < x1 + x2 + · · · + xn = n < n + 1 ≤ 1r , we get xn < 1r . Therefore,
we have r < xi < 1r for all numbers xi .
−1
1
We will apply now Corollary 1.7 to the function f (u) = u−r
, u > r. We have f 0 (u) = (u−r)
2
and
1
x2
4rx + 2
g(x) = f 0
=
, 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 1.7, 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 3.17. 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 non-negative
n−en−1
numbers such that a1 + a2 + · · · + an = n.
(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
1
n−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].
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2006.
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