ON JENSEN TYPE INEQUALITIES WITH ORDERED
VARIABLES
Jensen Type Inequalities
With Ordered Variables
VASILE CÎRTOAJE
Department of Automation and Computers
University of Ploieşti
Bucureşti 39, Romania
EMail: vcirtoaje@upg-ploiesti.ro
Vasile Cîrtoaje
vol. 9, iss. 1, art. 19, 2008
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Received:
15 October, 2007
Accepted:
05 January, 2008
Communicated by:
J.E. Pečarić
2000 AMS Sub. Class.:
26D10, 26D20.
Key words:
Convex function, k-arithmetic ordered variables, k-geometric ordered variables,
Jensen’s inequality, Karamata’s inequality.
Abstract:
In this paper, we present some basic results concerning an extension of Jensen
type inequalities with ordered variables to functions with inflection points, and
then give several relevant applications of these results.
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Contents
1
Basic Results
3
2
Applications
13
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With Ordered Variables
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vol. 9, iss. 1, art. 19, 2008
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1.
Basic Results
An n-tuple of real numbers X = (x1 , x2 , . . . , xn ) is said to be increasingly ordered
if x1 ≤ x2 ≤ · · · ≤ xn . If x1 ≥ x2 ≥ · · · ≥ xn , then X is decreasingly ordered.
n
In addition, a set X = (x1 , x2 , . . . , xn ) with x1 +x2 +···+x
= s is said to be kn
arithmetic ordered if k of the numbers x1 , x2 , . . . , xn are smaller than or equal to s,
and the other n − k are greater than or equal to s. On the assumption that x1 ≤ x2 ≤
· · · ≤ xn , X is k-arithmetic ordered if
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With Ordered Variables
Vasile Cîrtoaje
x1 ≤ · · · ≤ xk ≤ s ≤ xk+1 ≤ · · · ≤ xn .
vol. 9, iss. 1, art. 19, 2008
It is easily seen that
X1 = (s − x1 + xk+1 , s − x2 + xk+2 , . . . , s − xn + xk )
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is a k-arithmetic ordered set if X is increasingly ordered, and is an (n−k)-arithmetic
ordered set if X is decreasingly ordered.
Similarly, an n-tuple of positive real numbers A = (a1 , a2 , . . . , an ) with
√
n
a1 a2 · · · an = r is said to be k-geometric ordered if k of the numbers a1 , a2 , . . . , an
are smaller than or equal to r, and the other n − k are greater than or equal to r. Notice that
ak+1 ak+2
ak
A1 =
,
,...,
a1
a2
an
is a k-geometric ordered set if A is increasingly ordered, and is an (n − k)-geometric
ordered set if A is decreasingly ordered.
Theorem 1.1. Let n ≥ 2 and 1 ≤ k ≤ n − 1 be natural numbers, and let f (u) be a
function on a real interval I, which is convex for u ≥ s, s ∈ I, and satisfies
f (x) + kf (y) ≥ (1 + k)f (s)
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for any x, y ∈ I such that x ≤ y and x + ky = (1 + k)s. If x1 , x2 , . . . , xn ∈ I such
that
x1 + x2 + · · · + xn
=S≥s
n
and at least n − k of x1 , x2 , . . . , xn are smaller than or equal to S, then
f (x1 ) + f (x2 ) + · · · + f (xn ) ≥ nf (S).
Proof. We will consider two cases: S = s and S > s.
A. Case S = s. Without loss of generality, assume that x1 ≤ x2 ≤ · · · ≤ xn . Since
x1 + x2 + · · · + xn = ns, and at least n − k of the numbers x1 , x2 , . . . , xn are smaller
than or equal to s, there exists an integer n−k ≤ i ≤ n−1 such that (x1 , x2 , . . . , xn )
is an i-arithmetic ordered set, i.e.
Jensen Type Inequalities
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Vasile Cîrtoaje
vol. 9, iss. 1, art. 19, 2008
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x1 ≤ · · · ≤ xi ≤ s ≤ xi+1 ≤ · · · ≤ xn .
By Jensen’s inequality for convex functions,
f (xi+1 ) + f (xi+2 ) + · · · + f (xn ) ≥ (n − i)f (z),
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where
xi+1 + xi+2 + · · · + xn
, z ≥ s, z ∈ I.
n−i
Thus, it suffices to prove that
z=
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f (x1 ) + · · · + f (xi ) + (n − i)f (z) ≥ nf (s).
Let y1 , y2 , . . . , yi ∈ I be defined by
x1 + ky1 = (1 + k)s, x2 + ky2 = (1 + k)s, . . . , xi + kyi = (1 + k)s.
We will show that z ≥ y1 ≥ y2 ≥ · · · ≥ yi ≥ s. Indeed, we have
y1 ≥ y2 ≥ · · · ≥ yi ,
s − xi
yi − s =
≥ 0,
k
and
ky1 = (1 + k)s − x1
= (1 + k − n)s + x2 + · · · + xn
≤ (k + i − n)s + xi+1 + · · · + xn
= (k + i − n)s + (n − i)z ≤ kz.
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With Ordered Variables
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vol. 9, iss. 1, art. 19, 2008
Since z ≥ y1 ≥ y2 ≥ · · · ≥ yi ≥ s implies y1 , y2 , . . . , yi ∈ I, by hypothesis we have
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f (x1 ) + kf (y1 ) ≥ (1 + k)f (s),
f (x2 ) + kf (y2 ) ≥ (1 + k)f (s),
.........
f (xi ) + kf (yi ) ≥ (1 + k)f (s).
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Adding all these inequalities, we get
f (x1 ) + f (x2 ) + · · · + f (xi ) + k[f (y1 ) + f (y2 ) + · · · + f (yi )] ≥ i(1 + k)f (s).
Consequently, it suffices to show that
pf (z) + (i − p)f (s) ≥ f (y1 ) + f (y2 ) + · · · + f (yi ),
where p = n−i
≤ 1. Let t = pz + (1 − p)s, s ≤ t ≤ z. Since the decreasingly
k
~
~i =
ordered vector Ai = (t, s, . . . , s) majorizes the decreasingly ordered vector B
(y1 , y2 , . . . , yi ), by Karamata’s inequality for convex functions we have
f (t) + (i − 1)f (s) ≥ f (y1 ) + f (y2 ) + · · · + f (yi ).
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Adding this inequality to Jensen’s inequality for the convex function
pf (z) + (1 − p)f (s) ≥ f (t),
the conclusion follows.
B. Case S > s. The function f (u) is convex for u ≥ S, u ∈ I. According to the
result from Case A, it suffices to show that
f (x) + kf (y) ≥ (1 + k)f (S),
Jensen Type Inequalities
With Ordered Variables
Vasile Cîrtoaje
for any x, y ∈ I such that x < S < y and x + ky = (1 + k)S.
For x ≥ s, this inequality follows by Jensen’s inequality for convex function.
For x < s, let z be defined by x + kz = (1 + k)s. Since k(z − s) = s − x > 0
and k(y − z) = (1 + k)(S − s) > 0, we have
x < s < z < y,
vol. 9, iss. 1, art. 19, 2008
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s < S < y.
Since x + kz = (1 + k)s and x < z, we have by hypothesis
f (x) + kf (z) ≥ (1 + k)f (s).
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Therefore, it suffices to show that
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k[f (y) − f (z)] ≥ (1 + k)[f (S) − f (s)],
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which is equivalent to
f (y) − f (z)
f (S) − f (s)
≥
.
y−z
S−s
This inequality is true if
f (y) − f (z)
f (y) − f (s)
f (S) − f (s)
≥
≥
.
y−z
y−s
S−s
Close
The left inequality and the right inequality can be reduced to Jensen’s inequalities
for convex functions,
(y − z)f (s) + (z − s)f (y) ≥ (y − s)f (z)
and
(S − s)f (y) + (y − S)f (s) ≥ (y − s)f (S),
respectively.
Jensen Type Inequalities
With Ordered Variables
Remark 1. In the particular case k = n − 1, if f (x) + (n − 1)f (y) ≥ nf (s) for any
x, y ∈ I such that x ≤ y and x + (n − 1)y = ns, then the inequality in Theorem 1.1,
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vol. 9, iss. 1, art. 19, 2008
f (x1 ) + f (x2 ) + · · · + f (xn ) ≥ nf (S),
n
holds for any x1 , x2 , . . . , xn ∈ I which satisfy x1 +x2 +···+x
= S ≥ s. This result has
n
been established in [1, p. 143] and [2].
Remark 2. In the particular case k = 1 (when n − 1 of x1 , x2 , . . . , xn are smaller
than or equal to S), the hypothesis f (x) + kf (y) ≥ (1 + k)f (s) in Theorem 1.1 has
a symmetric form:
f (x) + f (y) ≥ 2f (s)
for any x, y ∈ I such that x + y = 2s.
(s)
Remark 3. Let g(u) = f (u)−f
. In some applications it is useful to replace the
u−s
hypothesis f (x) + kf (y) ≥ (1 + k)f (s) in Theorem 1.1 by the equivalent condition:
g(x) ≤ g(y) for any x, y ∈ I such that x < s < y and x + ky = (1 + k)s.
Their equivalence follows from the following observation:
f (x) + kf (y) − (1 + k)f (s) = f (x) − f (s) + k(f (y) − f (s))
= (x − s)g(x) + k(y − s)g(y)
= (x − s)(g(x) − g(y)).
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Remark 4. If f is differentiable on I, then Theorem 1.1 holds true by replacing the
hypothesis f (x) + kf (y) ≥ (1 + k)f (s) with the more restrictive condition:
f 0 (x) ≤ f 0 (y) for any x, y ∈ I such that x ≤ s ≤ y and x + ky = (1 + k)s.
To prove this assertion, we have to show that this condition implies f (x) + kf (y) ≥
(1 + k)f (s) for any x, y ∈ I such that x ≤ s ≤ y and x + ky = (1 + k)s. Let us
denote
s + ks − x
F (x) = f (x) + kf (y) − (1 + k)f (s) = f (x) + kf
− (1 + k)f (s).
k
0
0
Jensen Type Inequalities
With Ordered Variables
Vasile Cîrtoaje
vol. 9, iss. 1, art. 19, 2008
0
Since F (x) = f (x) − f (y) ≤ 0, F (x) is decreasing for x ∈ I, x ≤ s, and hence
F (x) ≥ F (s) = 0.
Remark 5. The inequality in Theorem 1.1 becomes equality for x1 = x2 = · · · =
xn = S. In the particular case S = s, if there are x, y ∈ I such that x < s < y,
x + ky = (k + 1)s and f (x) + kf (y) = (1 + k)f (s), then equality holds again for
x1 = x, x2 = · · · = xn−k = s and xn−k+1 = · · · = xn = y.
Remark 6. Let i be an integer such that n − k ≤ i ≤ n − 1. We may rewrite the
inequality in Theorem 1.1 as either
f (S − a1 + an−i+1 ) + f (S − a2 + an−i+2 ) + · · · + f (S − an + an−i ) ≥ nf (S)
with a1 ≥ a2 ≥ · · · ≥ an , or
f (S − a1 + ai+1 ) + f (S − a2 + ai+2 ) + · · · + f (S − an + ai ) ≥ nf (S)
with a1 ≤ a2 ≤ · · · ≤ an .
Corollary 1.2. Let n ≥ 2 and 1 ≤ k ≤ n − 1 be natural numbers, and let g be a
function on (0, ∞) such that f (u) = g(eu ) is convex for u ≥ 0, and
g(x) + kg(y) ≥ (1 + k)g(1)
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for any positive real numbers x and y with x ≤ y and xy k = 1. If a1 , a2 , . . . , an
√
are positive real numbers such that n a1 a2 · · · an = r ≥ 1 and at least n − k of
a1 , a2 , . . . , an are smaller than or equal to r, then
g(a1 ) + g(a2 ) + · · · + g(an ) ≥ ng(r).
Proof. We apply Theorem 1.1 to the function f (u) = g(eu ). In addition, we set
s = 0, S = ln r, and replace x with ln x, y with ln y, and each xi with ln ai .
Remark 7. If f is differentiable on (0, ∞), then Corollary 1.2 holds true by replacing
the hypothesis g(x) + kg(y) ≥ (1 + k)g(1) with the more restrictive condition:
Jensen Type Inequalities
With Ordered Variables
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vol. 9, iss. 1, art. 19, 2008
xg 0 (x) ≤ yg 0 (y) for all x, y > 0 such that x ≤ 1 ≤ y and xy k = 1.
To prove this claim, it suffices to show that this condition implies g(x) + kg(y) ≥
(1 + k)g(1) for all x, y > 0 with x ≤ 1 ≤ y and xy k = 1. Let us define the function
G by
r !
k 1
G(x) = g(x) + kg(y) − (1 + k)g(1) = g(x) + kg
− (1 + k)g(1).
x
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Since
1 0
xg 0 (x) − yg 0 (y)
G0 (x) = g 0 (x) − √
≤ 0,
g
(y)
=
x
xkx
G(x) is decreasing for x ≤ 1. Therefore, G(x) ≥ G(1) = 0 for x ≤ 1, and hence
g(x) + kg(y) ≥ (1 + k)g(1).
Remark 8. Let i be an integer such that n − k ≤ i ≤ n − 1. We may rewrite the
inequality for r = 1 in Corollary 1.2 as either
xn−i+1
xn−i+2
xn−i
g
+g
+ ··· + g
≥ ng(1)
x1
x2
xn
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for x1 ≥ x2 ≥ · · · ≥ xn > 0, or
xi+1
xi+2
xi
g
+g
+ ··· + g
≥ ng(1)
x1
x2
xn
for 0 < x1 ≤ x2 ≤ · · · ≤ xn .
Theorem 1.3. Let n ≥ 2 and 1 ≤ k ≤ n − 1 be natural numbers, and let f (u) be a
function on a real interval I, which is concave for u ≤ s, s ∈ I, and satisfies
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kf (x) + f (y) ≤ (k + 1)f (s)
vol. 9, iss. 1, art. 19, 2008
for any x, y ∈ I such that x ≤ y and kx + y = (k + 1)s. If x1 , x2 , . . . , xn ∈ I such
n
that x1 +x2 +···+x
= S ≤ s and at least n − k of x1 , x2 , . . . , xn are greater than or
n
equal to S, then
f (x1 ) + f (x2 ) + · · · + f (xn ) ≤ nf (S).
Proof. This theorem follows from Theorem 1.1 by replacing f (u) by −f (−u), s by
−s, S by −S, x by −y, y by −x, and each xi by −xn−i+1 for all i.
Remark 9. In the particular case k = n − 1, if (n − 1)f (x) + f (y) ≤ nf (s) for any
x, y ∈ I such that x ≤ y and (n − 1)x + y = ns, then the inequality in Theorem 1.3,
f (x1 ) + f (x2 ) + · · · + f (xn ) ≤ nf (S),
holds for any x1 , x2 , . . . , xn ∈ I which satisfy
been established in [1, p. 147] and [2].
x1 +x2 +···+xn
n
= S ≤ s. This result has
Remark 10. In the particular case k = 1 (when n − 1 of x1 , x2 , . . . , xn are greater
than or equal to S), the hypothesis kf (x) + f (y) ≤ (k + 1)f (s) in Theorem 1.3 has
a symmetric form: f (x) + f (y) ≤ 2f (s) for any x, y ∈ I such that x + y = 2s.
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(s)
Remark 11. Let g(u) = f (u)−f
. The hypothesis kf (x) + f (y) ≤ (k + 1)f (s) in
u−s
Theorem 1.3 is equivalent to
g(x) ≥ g(y) for any x, y ∈ I such that x < s < y and kx + y = (k + 1)s.
Remark 12. If f is differentiable on I, then Theorem 1.3 holds true if we replace the
hypothesis kf (x) + f (y) ≤ (k + 1)f (s) with the more restrictive condition
f 0 (x) ≥ f 0 (y) for any x, y ∈ I such that x ≤ s ≤ y and kx + y = (k + 1)s.
Remark 13. The inequality in Theorem 1.3 becomes equality for x1 = x2 = · · · =
xn = S. In the particular case S = s, if there are x, y ∈ I such that x < s < y,
kx + y = (k + 1)s and kf (x) + f (y) = (1 + k)f (s), then equality holds again for
x1 = · · · = xk = x, xk+1 = · · · = xn−1 = s and xn = y.
Remark 14. Let i be an integer such that 1 ≤ i ≤ k. We may rewrite the inequality
in Theorem 1.3 as either
f (S − a1 + ai+1 ) + f (S − a2 + ai+2 ) + · · · + f (S − an + ai ) ≤ nf (S)
with a1 ≤ a2 ≤ · · · ≤ an , or
f (S − a1 + an−i+1 ) + f (S − a2 + an−i+2 ) + · · · + f (S − an + an−i ) ≤ nf (S)
with a1 ≥ a2 ≥ · · · ≥ an .
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Corollary 1.4. Let n ≥ 2 and 1 ≤ k ≤ n − 1 be natural numbers, and let g be a
function on (0, ∞) such that f (u) = g(eu ) is concave for u ≤ 0, and
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kg(x) + g(y) ≤ (k + 1)g(1)
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for any positive real numbers x and y with x ≤ y and xk y = 1. If a1 , a2 , . . . , an
√
are positive real numbers such that n a1 a2 · · · an = r ≤ 1 and at least n − k of
a1 , a2 , . . . , an are greater than or equal to r, then
g(a1 ) + g(a2 ) + · · · + g(an ) ≤ ng(r).
Proof. We apply Theorem 1.3 to the function f (u) = g(eu ). In addition, we set
s = 0, S = ln r, and replace x with ln x, y with ln y, and each xi with ln ai .
Remark 15. If f is differentiable on (0, ∞), then Corollary 1.4 holds true by replacing the hypothesis kg(x) + g(y) ≤ (k + 1)g(1) with the more restrictive condition:
xg 0 (x) ≥ yg 0 (y) for all x, y > 0 such that x ≤ 1 ≤ y and xk y = 1.
Remark 16. Let i be an integer such that 1 ≤ i ≤ k. We may rewrite the inequality
for r = 1 in Corollary 1.4 as either
xi+1
xi+2
xi
g
+g
+ ··· + g
≤ ng(1)
x1
x2
xn
for 0 < x1 ≤ x2 ≤ · · · ≤ xn , or
xn−i+2
xn−i+1
xn−i
g
+g
+ ··· + g
≤ ng(1)
x1
x2
xn
for x1 ≥ x2 ≥ · · · ≥ xn > 0.
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vol. 9, iss. 1, art. 19, 2008
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2.
Applications
Proposition 2.1. Let n ≥ 2 and 1 ≤ k ≤ n − 1 be natural numbers, and let
x1 , x2 , . . . , xn be nonnegative real numbers such that x1 + x2 + · · · + xn = n.
(a) If at least n − k of x1 , x2 , . . . , xn are smaller than or equal to 1, then
k(x31 + x32 + · · · + x3n ) + (1 + k)n ≥ (1 + 2k)(x21 + x22 + · · · + x2n );
(b) If at least n − k of x1 , x2 , . . . , xn are greater than or equal to 1, then
x31
+
x32
+ ··· +
x3n
+ (k + 1)n ≤ (k +
2)(x21
+
x22
+ ··· +
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vol. 9, iss. 1, art. 19, 2008
x2n ).
Proof. (a) The inequality is equivalent to f (x1 ) + f (x2 ) + · · · + f (xn ) ≥ nf (S),
n
where S = x1 +x2 +···+x
= 1 and f (u) = ku3 − (1 + 2k)u2 . For u ≥ 1,
n
f 00 (u) = 2(3ku − 1 − 2k) ≥ 2(k − 1) ≥ 0.
Therefore, f is convex for u ≥ s = 1. According to Theorem 1.1 and Remark 3, we
have to show that g(x) ≤ g(y) for any nonnegative real numbers x < y such that
x + ky = 1 + k, where
g(u) =
Jensen Type Inequalities
With Ordered Variables
f (u) − f (1)
= ku2 − (1 + k)u − 1 − k.
u−1
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Indeed,
g(y) − g(x) = (k − 1)x(y − x) ≥ 0.
Equality occurs for x1 = x2 = · · · = xn = 1. On the assumption that x1 ≤
x2 ≤ · · · ≤ xn , equality holds again for x1 = 0, x2 = · · · = xn−k = 1 and
xn−k+1 = · · · = xn = 1 + k1 .
(b) Write the inequality as f (x1 ) + f (x2 ) + · · · + f (xn ) ≤ nf (S), where S =
x1 +x2 +···+xn
= 1 and f (u) = u3 − (k + 2)u2 . From the second derivative,
n
f 00 (u) = 2(3u − k − 2),
it follows that f is concave for u ≤ s = 1. According to Theorem 1.3 and Remark
11, we have to show that g(x) ≥ g(y) for any nonnegative real numbers x < y such
that kx + y = k + 1, where
g(u) =
f (u) − f (1)
= u2 − (k + 1)u − k − 1.
u−1
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It is easy to see that
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g(x) − g(y) = (k − 1)x(y − x) ≥ 0.
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Equality occurs for x1 = x2 = · · · = xn = 1. On the assumption that x1 ≤ x2 ≤
· · · ≤ xn , equality holds again for x1 = · · · = xk = 0, xk+1 = · · · = xn−1 = 1 and
xn = k + 1.
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Remark 17. For k = n − 1, the inequalities above become as follows
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(n − 1)(x31 + x32 + · · · + x3n ) + n2 ≥ (2n − 1)(x21 + x22 + · · · + x2n )
and
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x31
+
x32
+ ··· +
x3n
2
+ n ≤ (n +
1)(x21
+
x22
+ ··· +
x2n ),
respectively. By Remark 1 and Remark 9, these inequalities hold for any nonnegative
real numbers x1 , x2 , . . . , xn which satisfy x1 + x2 + · · · + xn = n (Problems 3.4.1
and 3.4.2 from [1, p. 154]).
Remark 18. For k = 1, we get the following statement:
Let x1 , x2 , . . . , xn be nonnegative real numbers such that x1 + x2 + · · · + xn = n.
Close
(a) If x1 ≤ · · · ≤ xn−1 ≤ 1 ≤ xn , then
x31 + x32 + · · · + x3n + 2n ≥ 3(x21 + x22 + · · · + x2n );
(b) If x1 ≤ 1 ≤ x2 ≤ · · · ≤ xn , then
x31 + x32 + · · · + x3n + 2n ≤ 3(x21 + x22 + · · · + x2n ).
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With Ordered Variables
Proposition 2.2. Let n ≥ 2 and 1 ≤ k ≤ n − 1 be natural numbers, and let
x1 , x2 , . . . , xn be positive real numbers such that x1 + x2 + · · · + xn = n. If at least
n − k of x1 , x2 , . . . , xn are greater than or equal to 1, then
1
1
1
4k
+
+ ··· +
−n≥
(x2 + x22 + · · · + x2n − n).
x1 x2
xn
(k + 1)2 1
Proof. Rewrite the inequality as f (x1 ) + f (x2 ) + · · · + f (xn ) ≤ nf (S), where
1
4ku2
n
S = x1 +x2 +···+x
= 1 and f (u) = (k+1)
2 − u . For 0 < u ≤ s = 1, we have
n
f 00 (u) =
Vasile Cîrtoaje
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2
8k
2
8k
−2(k − 1)
− 3 ≤
−2=
≤ 0;
2
2
(k + 1)
u
(k + 1)
(k + 1)2
therefore, f is concave on (0, 1]. By Theorem 1.3 and Remark 11, we have to show
that g(x) ≥ g(y) for any positive real numbers x < y such that kx + y = k + 1,
where
4k(u + 1) 1
f (u) − f (1)
=
+ .
g(u) =
u−1
(k + 1)2
u
Indeed,
1
4k
(y − x)(2kx − k − 1)2
g(x) − g(y) = (y − x)
−
=
≥ 0.
xy (k + 1)2
(k + 1)2 xy
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Equality occurs for x1 = x2 = · · · = xn = 1. Under the assumption that x1 ≤ x2 ≤
· · · ≤ xn , equality holds again for x1 = · · · = xk = k+1
, xk+1 = · · · = xn−1 = 1
2k
k+1
and xn = 2 .
Remark 19. For k = n − 1, the inequality in Proposition 2.2 becomes as follows:
1
1
1
4(n − 1) 2
+
+ ··· +
−n≥
(x1 + x22 + · · · + x2n − n).
x1 x2
xn
n2
By Remark 9, this inequality holds for any positive real numbers x1 , x2 , . . . , xn
which satisfy x1 + x2 + · · · + xn = n (Problems 3.4.5 from [1, p. 158]).
Remark 20. For k = 1, the following nice statement follows:
If x1 , x2 , . . . , xn are positive real numbers such that x1 ≤ 1 ≤ x2 ≤ · · · ≤ xn and
x1 + x2 + · · · + xn = n, then
1
1
1
+
+ ··· +
≥ x21 + x22 + · · · + x2n .
x1 x2
xn
Proposition 2.3. Let n ≥ 2 and 1 ≤ k ≤ n − 1 be natural numbers, and let
x1 , x2 , . . . , xn be nonnegative real numbers such that x1 + x2 + · · · + xn = n.
(a) If at least n − k of x1 , x2 , . . . , xn are smaller than or equal to 1, then
1
1
1
n
+
+ ··· +
≥
;
2
2
2
k + 1 + kx1 k + 1 + kx2
k + 1 + kxn
2k + 1
(b) If at least n − k of x1 , x2 , . . . , xn are greater than or equal to 1, then
k2
1
1
1
n
+ 2
+···+ 2
≤
.
2
2
2
+ k + 1 + kx1 k + k + 1 + kx2
k + k + 1 + kxn
(k + 1)2
Jensen Type Inequalities
With Ordered Variables
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Proof. (a) We may write the inequality as f (x1 ) + f (x2 ) + · · · + f (xn ) ≥ nf (S),
1
n
where S = x1 +x2 +···+x
= 1 and f (u) = k+1+ku
2 . Since the second derivative,
n
f 00 (u) =
2k(3ku2 − k − 1)
,
(k + 1 + ku2 )3
is positive for u ≥ 1, f is convex for u ≥ s = 1. According to Theorem 1.1 and
Remark 3, we have to show that g(x) ≤ g(y) for any nonnegative real numbers
x < y such that x + ky = 1 + k, where
Jensen Type Inequalities
With Ordered Variables
Vasile Cîrtoaje
vol. 9, iss. 1, art. 19, 2008
−k(u + 1)
f (u) − f (1)
g(u) =
=
.
u−1
(2k + 1)(k + 1 + ku2 )
Title Page
Indeed, we have
k 2 (y − x)
g(y) − g(x) =
(2k + 1)(k + 1 + kx2 )(k + 1 + ky 2 )
1
xy + x + y − 1 −
k
Contents
≥ 0,
since
1
x(2k − 1 + y)
=
≥ 0.
k
k
Equality occurs for x1 = x2 = · · · = xn = 1. On the assumption that x1 ≤
x2 ≤ · · · ≤ xn , equality holds again for x1 = 0, x2 = · · · = xn−k = 1 and
xn−k+1 = · · · = xn = 1 + k1 .
xy + x + y − 1 −
(b) We will apply Theorem 1.3 to the function f (u) =
Since the second derivative,
f 00 (u) =
1
k2 +k+1+ku2
2k(3ku2 − k 2 − k − 1)
,
(k 2 + k + 1 + ku2 )3
, for s = S = 1.
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is negative for 0 ≤ u < 1, f is concave for 0 ≤ u ≤ 1. According to Remark 11,
we have to show that g(x) ≥ g(y) for any nonnegative real numbers x < y such that
kx + y = k + 1, where
g(u) =
f (u) − f (1)
−k(u + 1)
=
.
2
u−1
(k + 1) (k 2 + k + 1 + ku2 )
Jensen Type Inequalities
With Ordered Variables
We have
k 2 (y − x)
g(x) − g(y) =
(k + 1)2 (k 2 + k + 1 + kx2 )(k 2 + k + 1 + ky 2 )
1
× k + + 1 − xy − x − y ≥ 0,
k
vol. 9, iss. 1, art. 19, 2008
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since
k+
1
1
+ 1 − xy − x − y = k x −
k
k
2
≥ 0.
Equality occurs for x1 = x2 = · · · = xn = 1. On the assumption that x1 ≤ x2 ≤
· · · ≤ xn , equality holds again for x1 = · · · = xk = k1 , xk+1 = · · · = xn−1 = 1 and
xn = k.
Remark 21. For k = n − 1, the inequalities in Proposition 2.3 become as follows:
1
1
1
n
+
+ ··· +
≥
2
2
2
n + (n − 1)x1 n + (n − 1)x2
n + (n − 1)xn
2n − 1
and
n2
Vasile Cîrtoaje
1
1
1
1
+ 2
+· · ·+ 2
≤ ,
2
2
2
− n + 1 + (n − 1)x1 n − n + 1 + (n − 1)x2
n − n + 1 + (n − 1)xn
n
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respectively. By Remark 1 and Remark 9, these inequalities hold for any nonnegative
numbers x1 , x2 , . . . , xn which satisfy x1 + x2 + · · · + xn = n (Problems 3.4.3 and
3.4.4 from [1, p. 156]).
Remark 22. For k = 1, we get the following statement:
Let x1 , x2 , . . . , xn be nonnegative real numbers such that x1 + x2 + · · · + xn = n.
(a) If x1 ≤ · · · ≤ xn−1 ≤ 1 ≤ xn , then
1
1
1
n
+
+ ··· +
≥
;
2
2
2 + x1 2 + x2
2 + x2n
3
Jensen Type Inequalities
With Ordered Variables
Vasile Cîrtoaje
vol. 9, iss. 1, art. 19, 2008
(b) If x1 ≤ 1 ≤ x2 ≤ · · · ≤ xn , then
1
1
1
n
+
+ ··· +
≤ .
2
2
2
3 + x1 3 + x2
3 + xn
4
Remark 23. By Theorem 1.1 and Theorem 1.3, the following more general statement
holds:
Let n ≥ 2 and 1 ≤ k ≤ n − 1 be natural numbers, and let x1 , x2 , . . . , xn be
nonnegative real numbers such that x1 + x2 + · · · + xn = nS.
(a) If S ≥ 1 and at least n − k of x1 , x2 , . . . , xn are smaller than or equal to S,
then
1
1
n
1
+
+ ··· +
≥
;
2
2
2
k + 1 + kx1 k + 1 + kx2
k + 1 + kxn
k + 1 + kS 2
(b) If S ≤ 1 and at least n − k of x1 , x2 , . . . , xn are greater than or equal to S, then
1
1
1
n
+
+·
·
·+
≤
.
2
2
k 2 + k + 1 + kx1 k 2 + k + 1 + kx2
k 2 + k + 1 + kx2n
k 2 + k + 1 + kS 2
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Proposition 2.4. Let n ≥ 2 and 1 ≤ k ≤ n − 1 be natural numbers, and let
a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1.
(a) If at least n − k of x1 , x2 , . . . , xn are smaller than or equal to 1, then
1
1
1
n
+
+ ··· +
≥
;
1 + ka1 1 + ka2
1 + kan
1+k
(b) If at least n − k of x1 , x2 , . . . , xn are greater than or equal to 1, then
1
1
n
1
+
+ ··· +
≤
.
a1 + k a2 + k
an + k
1+k
Proof. (a) We will apply Corollary 1.2 to the function g(x) =
1
function f (u) = g(eu ) = 1+ke
u has the second derivative
1
,
1+kx
Jensen Type Inequalities
With Ordered Variables
Vasile Cîrtoaje
vol. 9, iss. 1, art. 19, 2008
for r = 1. The
keu (keu − 1)
,
f (u) =
(1 + keu )3
00
which is positive for u > 0. Therefore, f is convex for u ≥ 0. Thus, it suffices to
show that g(x) + kg(y) ≥ (1 + k)g(1) for any x, y > 0 such that xy k = 1. The
inequality g(x) + kg(y) ≥ (1 + k)g(1) is equivalent to
yk
k
+
≥ 1,
k
y + k 1 + ky
or, equivalently,
y k + k − 1 ≥ ky.
The last inequality immediately follows from the AM-GM inequality applied to the
positive numbers y k , 1, . . . , 1. Equality occurs for a1 = a2 = · · · = an = 1.
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(b) We can obtain the required inequality either by replacing each number ai with its
reverse a1i in the inequality in part (a), or by means of Corollary 1.4. Equality occurs
for a1 = a2 = · · · = an = 1.
Remark 24. For k = n − 1, we get the known inequalities
1
1
1
+
+ ··· +
≥1
1 + (n − 1)a1 1 + (n − 1)a2
1 + (n − 1)an
and
1
1
1
+
+ ··· +
≤ 1,
a1 + n − 1 a2 + n − 1
an + n − 1
which hold for any positive numbers a1 , a2 , . . . , an such that a1 a2 · · · an = 1.
Remark 25. Using the substitution a1 = xk+1
, a2 = xk+2
, . . . , an = xxnk , we get the
x1
x2
following statement:
Let n ≥ 2 and 1 ≤ k ≤ n − 1 be natural numbers, and let x1 , x2 , . . . , xn be
positive real numbers.
(a) If x1 ≥ x2 ≥ · · · ≥ xn , then
x1
x2
xn
n
+
+ ··· +
≥
;
x1 + kxk+1 x2 + kxk+2
xn + kxk
1+k
(b) If x1 ≤ x2 ≤ · · · ≤ xn , then
x1
x2
xn
n
+
+ ··· +
≤
.
kx1 + xk+1 kx2 + xk+2
kxn + xk
k+1
In the particular case k = 1, we get
x1
x2
xn
n
+
+ ··· +
≥
x1 + x2 x2 + x3
xn + x1
2
Jensen Type Inequalities
With Ordered Variables
Vasile Cîrtoaje
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for x1 ≥ x2 ≥ · · · ≥ xn > 0, and
x1
x2
xn
n
+
+ ··· +
≤
x1 + x2 x2 + x3
xn + x1
2
for 0 < x1 ≤ x2 ≤ · · · ≤ xn .
Remark 26. By Corollary 1.2 and Corollary 1.4, we can see that the following more
general statement holds:
Let n ≥ 2 and 1 ≤ k ≤ n − 1 be natural numbers, and let a1 , a2 , . . . , an be
√
positive real numbers such that n a1 a2 · · · an = r.
(a) If r ≥ 1, and at least n − k of a1 , a2 , . . . , an are smaller than or equal to r, then
1
1
1
n
+
+ ··· +
≥
;
1 + ka1 1 + ka2
1 + kan
1 + kr
(b) If r ≤ 1, and at least n − k of a1 , a2 , . . . , an are greater than or equal to r, then
1
1
1
n
+
+ ··· +
≤
.
a1 + k a2 + k
an + k
r+k
Jensen Type Inequalities
With Ordered Variables
Vasile Cîrtoaje
vol. 9, iss. 1, art. 19, 2008
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Proposition 2.5. Let a1 , a2 , . . . , an be positive numbers such that a1 a2 · · · an = 1.
(a) If a1 ≤ · · · ≤ an−1 ≤ 1 ≤ an , then
√
1
1
1
n
+√
+ ··· + √
≥ ;
2
1 + 3a1
1 + 3a2
1 + 3an
(b) If a1 ≤ 1 ≤ a2 ≤ · · · ≤ an , then
√
1
1
1
n
+√
+ ··· + √
≤√ .
1 + 2a1
1 + 2a2
1 + 2an
3
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Proof. (a) We will apply Corollary 1.2 (case k = 1 and r = 1) to the function
1
1
g(x) = √1+3x
. The function f (u) = g(eu ) = √1+3e
u has the second derivative
5
1
f 00 (u) = eu (3eu − 2)(1 + 3eu )− 2 .
2
00
Since f > 0 for u ≥ 0, f is convex for u ≥ 0. Therefore, to finish the proof, we
have to show that g(x) + g(y) ≥ 2g(1) for any x, y > 0 with xy = 1. This inequality
is equivalent to
r
x
1
√
+
≥ 1.
x+3
1 + 3x
Using the substitution
√ 1
1+3x
= t, 0 < t < 1, transforms the inequality into
r
1 − t2
≥ 1 − t.
8t2 + 1
00
u
u
u − 52
f = e (e − 1)(1 + 2e )
vol. 9, iss. 1, art. 19, 2008
Contents
√ 1
.
1+2x
≤ 0.
Thus, it suffices to show that g(x) + g(y) ≤ 2g(1) for any x, y > 0 with xy = 1.
This inequality follows from the Cauchy-Schwarz inequality, as follows
s
r
r
3
3
3
3
+
≤
+1
1+
= 2.
1 + 2x
1 + 2y
1 + 2x
1 + 2y
Equality occurs for a1 = a2 = · · · = an = 1.
Vasile Cîrtoaje
Title Page
By squaring, we get t(1 − t)(2t − 1)2 ≥ 0, which is clearly true. Equality occurs for
a1 = a2 = · · · = an = 1.
(b) We will apply Corollary 1.4 (case k = 1 and r = 1) to the function g(x) =
1
The function f (u) = g(eu ) = √1+2e
u is concave for u ≤ 0, since
Jensen Type Inequalities
With Ordered Variables
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Remark 27. Using the substitution a1 = xx21 , a2 =
following statement:
Let x1 , x2 , . . . , xn be positive real numbers.
x3
,...,
x2
an =
x1
,
xn
we get the
(a) If x1 ≥ x2 ≥ · · · ≥ xn , then
r
r
r
xn
n
x2
x1
≥ ;
+
+ ··· +
xn + 3x1
2
x2 + 3x3
x1 + 3x2
(b) If x1 ≤ x2 ≤ · · · ≤ xn , then
r
r
r
3xn
3x2
3x1
≤ n.
+ ··· +
+
xn + 2x1
x2 + 2x3
x1 + 2x2
Remark 28. By Corollary 1.2 and Corollary 1.4, the following more general statement holds:
√
Let a1 , a2 , . . . , an be positive real numbers such that n a1 a2 · · · an = r.
(a) If r ≥ 1 and a1 ≤ · · · ≤ an−1 ≤ r ≤ an , then
√
1
1
1
n
+√
+ ··· + √
≥√
;
1 + 3a1
1 + 3a2
1 + 3an
1 + 3r
Jensen Type Inequalities
With Ordered Variables
Vasile Cîrtoaje
vol. 9, iss. 1, art. 19, 2008
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(b) If r ≤ 1 and a1 ≤ r ≤ a2 ≤ · · · ≤ an , then
1
1
1
n
√
+√
+ ··· + √
≤√
.
1 + 2a1
1 + 2a2
1 + 2an
1 + 2r
Proposition 2.6. Let a1 , a2 , . . . , an be positive numbers such that a1 a2 · · · an = 1.
Close
(a) If a1 ≤ · · · ≤ an−1 ≤ 1 ≤ an , then the following inequality holds for 0 ≤ p ≤
p0 , where p0 ∼
= 1.5214 is the positive root of the equation p3 − p − 2 = 0:
1
1
1
n
+
+
·
·
·
+
≥
;
(p + a1 )2 (p + a2 )2
(p + an )2
(p + 1)2
√
(b) If a1 ≤ 1 ≤ a2 ≤ · · · ≤ an , then the following inequality holds for p ≥ 1 + 2:
1
1
n
1
+
+ ··· +
≤
.
2
2
2
(p + a1 )
(p + a2 )
(p + an )
(p + 1)2
Proof. (a) We will apply Corollary 1.2 (case k = 1 and r = 1) to the function
1
1
u
g(x) = (p+x)
2 . Notice that the function f (u) = g(e ) = (p+eu )2 is convex for u ≥ 0,
because
2eu (2eu − p)
f 00 (u) =
> 0.
(p + eu )4
Consequently, we have to show that g(x) + g(y) ≥ 2g(1) for any x, y > 0 with
xy = 1; that is
1
2
1
+
≥
.
(p + x)2 (p + y)2
(p + 1)2
Using the substitution x + y = 2t, t ≥ 1, the inequality transforms into
2
Jensen Type Inequalities
With Ordered Variables
Vasile Cîrtoaje
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2
2t + 2pt + p − 1
1
≥
,
2
2
(2pt + p + 1)
(p + 1)2
or, equivalently,
(t − 1)[(1 + 2p − p2 )t + (1 − p)(p2 + 1)] ≥ 0.
Close
It is true, because 1 + 2p − p2 > p(2 − p) > 0 and
(1 + 2p − p2 )t + (1 − p)(p2 + 1) ≥ (1 + 2p − p2 ) + (1 − p)(p2 + 1)
= 2 + p − p3 ≥ 0
for 0 ≤ p ≤ p0 . Equality holds for a1 = a2 = · · · = an = 1.
(b) We will apply Corollary 1.4 (case k = 1 and r = 1) to the function g(x) =
The function f (u) = g(eu ) = (p+e1 u )2 is concave for u ≤ 0, since
1
.
(p+x)2
Jensen Type Inequalities
With Ordered Variables
Vasile Cîrtoaje
2eu (2eu − p)
f 00 (u) =
< 0.
(p + eu )4
By Corollary 1.4, it suffices to show that g(x) + g(y) ≤ 2g(1) for any x, y > 0 with
xy = 1; that is
1
1
2
+
≤
.
2
2
(p + x)
(p + y)
(p + 1)2
Using the notation x + y = 2t, t ≥ 1, the inequality becomes
(t − 1)[(p2 − 2p − 1)t + (p − 1)(p2 + 1)] ≥ 0.
√
It is true, since p2 − 2p − 1 ≥ 0 for p ≥ 1 + 2. Equality holds for a1 = a2 = · · · =
an = 1.
Remark 29. Using the substitution a1 = xx21 , a2 = xx32 , . . . , an = xxn1 , we get the
following statement:
Let x1 , x2 , . . . , xn be positive real numbers.
(a) If 0 ≤ p ≤ p0 ∼
= 1.5214 and x1 ≥ x2 ≥ · · · ≥ xn , then
2 2
2
x1
x2
xn
n
+
+ ··· +
≥
;
px1 + x2
px2 + x3
pxn + x1
(p + 1)2
vol. 9, iss. 1, art. 19, 2008
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(b) If p ≥ 1 +
√
2 and x1 ≤ x2 ≤ · · · ≤ xn , then
2 2
2
x1
x2
xn
n
+
+ ··· +
≤
.
px1 + x2
px2 + x3
pxn + x1
(p + 1)2
Remark 30. By Corollary 1.2 and Corollary 1.4, the following more general statement holds:
√
Let a1 , a2 , . . . , an be positive real numbers such that n a1 a2 · · · an = r.
Jensen Type Inequalities
With Ordered Variables
Vasile Cîrtoaje
(a) If r ≥ 1 and a1 ≤ · · · ≤ an−1 ≤ r ≤ an , then the following inequality
holds for 0 ≤ p ≤ p0 , where p0 ∼
= 1, 5214 is the positive root of the equation
3
p − p − 2 = 0:
vol. 9, iss. 1, art. 19, 2008
Title Page
1
1
1
n
+
+ ··· +
≥
;
2
2
2
(p + a1 )
(p + a2 )
(p + an )
(p + r)2
(b) If r ≤ 1 √
and a1 ≤ r ≤ a2 ≤ · · · ≤ an , then the following inequality holds for
p ≥ 1 + 2:
1
1
1
n
+
+ ··· +
≤
.
2
2
2
(p + a1 )
(p + a2 )
(p + an )
(p + r)2
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References
[1] V. CÎRTOAJE, A generalization of Jensen’s inequality, Gazeta Matematica A, 2
(2005), 124–138.
[2] V. CÎRTOAJE, Algebraic Inequalities - Old and New Methods, GIL Publishing
House, Romania, 2006.
[3] D.S. MITRINOVIĆ, J.E. PEČARIĆ
equalities in Analysis, Kluwer, 1993.
AND
[4] G.H. HARDY, J.E. LITTLEWOOD
University Press, 1952.
AND
A.M. FINK, Classical and New In-
Jensen Type Inequalities
With Ordered Variables
Vasile Cîrtoaje
vol. 9, iss. 1, art. 19, 2008
G. PÓLYA, Inequalities, Cambridge
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