INEQUALITIES VIA CONVEX FUNCTIONS

Internat. J. Math. & Math. Sci.
Vol. 22, No. 3 (1999) 543–546
S 0161-1712〈99〉22543-4
© Electronic Publishing House
INEQUALITIES VIA CONVEX FUNCTIONS
I. A. ABOU-TAIR and W. T. SULAIMAN
(Received 25 May 1998)
Abstract. A general inequality is proved using the definition of convex functions. Many
major inequalities are deduced as applications.
Keywords and phrases. Convex functions, inequalities.
1991 Mathematics Subject Classification. 26A51, 26D15.
1. Introduction. Kapur and Kumer (1986) have used the principle of dynamical
programming to prove major inequalities due to Shannon, Renyi, and Hölder. See [1].
In this note, we prove a general inequality using convex functions. As a result, the
inequalities of Shannon, Renyi, Hölder, and others are all deduced.
Let I be an interval in R, f : I → R is said to be convex if and only if, for all x, y ∈ I,
all λ, 0≤ λ ≤ 1,
f [ λx +(1−λ)y ] ≤ λf (x)+(1−λ)y. (1)
Here, we give the following new definitions:
(a) Let f and g be two functions and let I be an interval in R for which f ◦ g is
defined, then f is said to be g-convex if and only if, for all x, y ∈ I, all λ,
0 ≤ λ ≤ 1,
f [ λg(x)+(1−λ)g(y) ] ≤ λf ◦g(x)+(1−λ)f ◦g(y). (2)
(b) If the inequality is reversed, then f is said to be g-concave.
If g(x) = x, the two definitions of g-convex and convex functions become identical.
Theorem 1.1. Let f be g-convex, then
(i) if g is linear, then f ◦g is convex, and
(ii) if f is increasing and g is convex, then f ◦g is convex.
Proof.
(i)
(ii)
f ◦g [ λx +(1−λ)y ] = f [ λg(x)+(1−λ)g(y) ]
≤ λf ◦g(x)+(1−λ)f ◦g(y).
f ◦g [ λx +(1−λ)y ] ≤ f [ λg(x)+(1−λ)g(y) ]
≤ λf ◦g(x)+(1−λ)f ◦g(y).
(3)
(4)

544 I. A. ABOU-TAIR AND W. T. SULAIMAN
Lemma 1.1. Let f be g-convex and let ∑ n
i=1 t i = T n = 1, t i ≥ 0, i = 1,2,...,n, then
(
∑ n
f t i g ( ) ) n∑
x i ≤ t i f ◦g ( )
x i . (5)
Proof.
(
∑ n
f
i=1
i=1
t i g ( ) ) n−1
∑
x i =f
(T n−1
≤T n−1 f
=T n−2 f
i=1
i=1
t i
g ( )
x i +tn g ( ) )
x n
T n−1
( n−1 ∑ t i
g ( ) )
x i +t n f ◦g ( )
x n
T
i=1 n−1
(
n−2
Tn−2 ∑
T n−1
i=1
( n−2 ∑ t i
i=1
t i
g ( ) t n−1
x i + g ( ) )
x n−1 +t n f ◦g ( )
x n
T n−2 T n−1
(6)
≤T n−2 f
g ( ) )
x i +t n−1 f ◦g ( )
x n−1 +tn f ◦g ( )
x n
T n−2 .
n∑
≤ t i f ◦g ( )
x i .
i=1
Lemma 1.2. For any function g, the exponential function f(x)= e x is g-convex.
Proof.
Define
F(x)= λe g(x) +(1−λ)e g(y) −e λg(x)+(1−λ)g(y) . (7)
Let
G(t) = (1−λ)+λt −t λ , t >0. (8)
It follows that
G ′ (t) = λ ( 1−t λ−1) , G ′′ (t) = λ(1−λ)t λ−2 . (9)
Thus, G ′ (t) = 0 when t = 1 and G ′′ (1) = λ(1 − λ) > 0. Hence, G has its minimum
value 0 at t = 1 and this implies G(t) ≥ 0, t>0. The result follows by putting F(x)=
e g(y) G(e g(x)−g(y) ).
Corollary 1.3. The function f(x) = ln(x) is concave for if h(x) = e x , then, by
Lemma 1.2, h is f -convex. Hence,
It follows that
e λ(lnx)+(1−λ)lny ≤ λe lnx +(1−λ)e lny = λx +(1−λy). (10)
λlnx +(1−λ)lny ≤ ln [ λx +(1−λ)y ] . (11)
2. Main inequality
Theorem 2.1.
Proof.
n∑
m∏
j=1 i=1
(
pij
) qi
/ ∑mi=1
q i
≤
∑ m
i=1
∑ n
j=1 p ij q i
∑ m
i=1 q i
. (12)
If f(x)= e x and g(x) = lnx, then f is g-convex. By Lemma 1.2, we have

INEQUALITIES VIA CONVEX FUNCTIONS 545
m∏
/
( ) ∑mi=1 (
/ ∑mi=1 ∏mi=1
)
qi q i pij = e ln (p ij ) q i q i
Therefore,
i=1
j=1 i=1
/
∑ ∑mi=1
mi=1
ln(p
= e
ij ) q i q ∑ i mi=1
/ ∑mi=1
(q
= e i q i )lnp ij
( ∑ m∑
m
qi
≤ ∑ m
)e lnp ij i=1 q i p ij
= ∑ m .
i=1 q i i=1 q i
i=1
(13)
n∑ m∏
/
( ) ∑mi=1
∑ n ∑ m ∑ m ∑ n
qi q i j=1 i=1 p ij q i i=1 j=1 p ij q i
pij ≤ ∑ m = ∑ m . (14)
i=1 q i i=1 q i
3. Applications
Theorem 3.1 (Shannon’s inequality). Given ∑ m
i=1 a i = a, ∑ m
m∑
Proof.
we have
That is
It follows that
Hence, we get
( ) a
aln ≤
b
i=1
Applying Theorem 2.1 by putting
p ij = b i
a i
, j = 1, q i = a i ,
m∏
i=1
i=1 b i = b, then
( )
ai
a i ln , a i ,b i ≥ 0. (15)
b i
m∑
a i = a,
i=1
m∑
b i = b, (16)
/
( ) ∑mi=1
ai a bi
i
∑ m
i=1 b
≤ ∑ i
m . (17)
a i i=1 a i
m∏
i=1
i=1
(
bi
a i
) ai /a
≤ b a . (18)
a
m
b ≤ ∏ ( ) ai /a ai
. (19)
b i
i=1
( ) a
aln ≤
b
m∑
i=1
( )
ai
a i ln . (20)
b i
Theorem 3.2 (Renyi’s inequality). Given ∑ m
i=1 a i = a, ∑ m
i=1 b i = b, then, for α>0,
α ̸=1,
1 (
a α b 1−α −a ) m∑ 1 ( )
≤ a
α
i
α−1
α−1
b1−α i
−a i , ai ,b i ≥ 0. (21)
i=1
Proof. Applying Theorem 2.1 with i = 2, p 1j = c j , p 2j = d j , q 1 = λ, q 2 = 1 − λ,
0

546 I. A. ABOU-TAIR AND W. T. SULAIMAN
/∑ m
/∑ m
On putting c j = (a j j=1 a j ) and d j = (b j j=1 b j ), inequality (22) implies
( m∑
m
)
∑ λ ( m
)
∑ 1−λ
a λ j b1−λ j
≤ a j b j , (23)
and this gives
j=1
j=1
a λ b 1−λ
λ−1 ≤ 1
λ−1
m∑
j=1
j=1
a λ j b1−λ j
. (24)
Thus, for the case 0

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