# An extension of Jensen's discrete inequality to half convex functions

## Abstract

We extend the right and left convex function theorems to weighted Jensen's type inequalities, and then combine the new theorems in a single one applicable to a half convex function f(u), defined on a real interval $I$ and convex for us or us, where $s∈I$. The obtained results are applied for proving some open relevant inequalities.

## 1 Introduction

The right convex function theorem (RCF-Theorem) has the following statement (see ).

RCF-Theorem. Let f(u) be a function defined on a real interval $I$ and convex for $u ≥ s ∈ I$. The inequality

$f ( x 1 ) +f ( x 2 ) +⋯+f ( x n ) ≥nf ( x 1 + x 2 + ⋯ + x n n )$
(1)

holds for all$x 1 , x 2 ,…, x n ∈I$satisfying x1 + x2 + ... + x n ns if and only if

$f ( x ) + ( n - 1 ) f ( y ) ≥nf ( s )$
(2)

for all$x,y∈I$which satisfy xsy and x + (n - 1)y = ns.

Replacing f(u) by f(-u), s by -s, x by -x, y by -y, and each x i by -x i for i = 1, 2, ..., n, from RCF-Theorem we get the left convex function theorem (LCF-Theorem).

LCF-Theorem. Let f(u) be a function defined on a real interval $I$ and convex for $u≤s∈I$. The inequality

$f ( x 1 ) + f ( x 2 ) + ⋯ + f ( x n ) ≥ n f ( x 1 + x 2 + ⋯ + x n n )$
(3)

holds for all$x 1 , x 2 ,…, x n ∈I$satisfying x1 + x2 + ... + x n ns if and only if

$f ( x ) + ( n - 1 ) f ( y ) ≥nf ( s )$
(4)

for all$x,y∈I$which satisfy xsy and × + (n - 1)y = ns.

Notice that from RCF- and LCF-Theorems, we get the following theorem, which we have called the half convex function theorem (HCF-Theorem).

HCF-Theorem. Let f(u) be a function defined on a real interval $I$ and convex for us or us, where $s∈I$. The inequality

$f ( x 1 ) +f ( x 2 ) +⋯+f ( x n ) ≥nf ( x 1 + x 2 + ⋯ + x n n )$
(5)

holds for all$x 1 , x 2 ,…, x n ∈I$satisfying x1 + x2 + ... + x n = ns if and only if

$f ( x ) + ( n - 1 ) f ( y ) ≥nf ( s )$
(6)

for all$x,y∈I$which satisfy x + (n - 1)y = ns.

Applying RCF-, LCF-, and HCF-Theorems to the function f(u) = g(eu ), and replacing s by ln r, x by ln x, y by ln y, and each x i by ln a i for i = 1, 2, ..., n, we get the following corollaries, respectively.

RCF-Corollary. Let g be a function defined on a positive interval $I$ such that f(u) = g(eu ) is convex for $e u ≥r∈I$. The inequality

$g ( a 1 ) + g ( a 2 ) + ⋯ + g ( a n ) ≥ n g ( a 1 a 2 … a n n )$
(7)

holds for all$a 1 , a 2 ,…, a n ∈I$satisfying a1a2 ... a n rn if and only if

$g ( a ) + ( n - 1 ) g ( b ) ≥ n g ( r )$
(8)

for all$a,b∈I$which satisfy arb and abn-1= rn .

LCF-Corollary. Let g be a function defined on a positive interval $I$ such that f(u) = g(eu ) is convex for $e u ≤r∈I$. The inequality

$g ( a 1 ) + g ( a 2 ) + ⋯ + g ( a n ) ≥ n g ( a 1 a 2 … a n n )$
(9)

holds for all$a 1 , a 2 ,…, a n ∈I$satisfying a1a2 ... a n rn if and only if

$g ( a ) + ( n - 1 ) g ( b ) ≥ng ( r )$
(10)

for all$a,b∈I$which satisfy arb and abn-1= rn .

HCF-Corollary. Let g be a function defined on a positive interval $I$ such that f(u) = g(eu ) is convex for eur or eur, where $r∈I$. The inequality

$g ( a 1 ) +g ( a 2 ) +⋯+g ( a n ) ≥ng ( r )$
(11)

holds for all$a 1 , a 2 ,…, a n ∈I$satisfying a1a2 ... a n = rn if and only if

$g ( a ) + ( n - 1 ) g ( b ) ≥ng ( r )$
(12)

for all$a,b∈I$which satisfy abn-1= rn .

## 2 Main results

In order to extend RCF-, LCF-, and HCF-Theorems to weighted Jensen's type inequalities, we need the following lemma.

Lemma 2.1 Let q1, q2and r1, r2, ..., r m be nonnegative real numbers such that

$r 1 + r 2 +⋯+ r m = q 1 + q 2 ,$
(13)

and let f be a convex function on$I$. If$a,b∈I$(ab) and x1, x2, ..., x m [a, b] such that

$r 1 x 1 + r 2 x 2 +⋯+ r m x m = q 1 a+ q 2 b,$
(14)

then

$r 1 f ( x 1 ) + r 2 f ( x 2 ) +⋯+ r m f ( x m ) ≤ q 1 f ( a ) + q 2 f ( b ) .$
(15)

The weighted right convex function theorem (WRCF-Theorem), weighted left convex function theorem (WLCF-Theorem), and weighted half convex function theorem (WHCF-Theorem) are the following.

WRCF-Theorem. Let f(u) be a function defined on a real interval $I$ and convex for $u≥s∈I$, and let p1, p2, ..., p n be positive real numbers such that

$p=min { p 1 , p 2 , … , p n } , p 1 + p 2 +⋯+ p n =1.$
(16)

The inequality

$p 1 f ( x 1 ) + p 2 f ( x 2 ) +⋯+ p n f ( x n ) ≥f ( p 1 x 1 + p 2 x 2 + ⋯ + p n x n )$
(17)

holds for all$x 1 , x 2 ,…, x n ∈I$satisfying p1x1 + p2x2 + ... + p n x n s if and only if

$pf ( x ) + ( 1 - p ) f ( y ) ≥f ( s )$
(18)

for all$x,y∈I$such that xsy and px + (1 - p)y = s.

WLCF-Theorem. Let f(u) be a function defined on a real interval $I$ and convex for $u≤s∈I$, and let p1, p2 , ..., p n be positive real numbers such that

$p=min { p 1 , p 2 , … , p n } , p 1 + p 2 +⋯+ p n =1.$
(19)

The inequality

$p 1 f ( x 1 ) + p 2 f ( x 2 ) +⋯+ p n f ( x n ) ≥f ( p 1 x 1 + p 2 x 2 + ⋯ + p n x n )$
(20)

holds for all$x 1 , x 2 ,…, x n ∈I$satisfying p1x1 + p2x2 + ... + p n x n s if and only if

$pf ( x ) + ( 1 - p ) f ( y ) ≥f ( s )$
(21)

for all$x,y∈I$such that xsy and px + (1 - p)y = s.

WHCF-Theorem. Let f(u) be a function defined on a real interval $I$ and convex for us or us, where $s∈I$, and let p1, p2, ..., p n be positive real numbers such that

$p=min { p 1 , p 2 , … , p n } , p 1 + p 2 +⋯+ p n =1.$
(22)

The inequality

$p 1 f ( x 1 ) + p 2 f ( x 2 ) +⋯+ p n f ( x n ) ≥f ( p 1 x 1 + p 2 x 2 + ⋯ + p n x n )$
(23)

holds for all$x 1 , x 2 ,…, x n ∈I$satisfying p1x1 + p2x2 + ... + p n x n = s if and only if

$pf ( x ) + ( 1 - p ) f ( y ) ≥f ( s )$
(24)

for all$x,y∈I$such that px + (1 - p)y = s.

Notice that WLCF-Theorem can be obtained from WRCF-Theorem replacing f(u) by f(-u), s by -s, x by -x, y by -y, and each x i by -x i for i = 1,2, ..., n.

On the other hand, applying WRCF-, WLCF-, and WHCF-Theorems to the function f(u) = g(eu ) and replacing s by 1nr, x by 1nx, y by 1ny, and each x i by In a i for i = 1, 2, ... n, we get the following corollaries, respectively.

WRCF-Corollary Let g be a function defined on a positive interval $I$ such that f(u) = g(eu ) is convex for $lnu≥r∈I$, and let p1, p2, ..., p n be positive real numbers such that

$p=min { p 1 , p 2 , … , p n } , p 1 + p 2 +⋯+ p n =1.$
(25)

The inequality

$p 1 g ( a 1 ) + p 2 g ( a 2 ) +⋯+ p n g ( a n ) ≥g ( a 1 p 1 a 2 p 2 … a n p n )$
(26)

holds for all $a 1 , a 2 ,…, a n ∈I$ satisfying $a 1 p 1 a 2 p 2 … a n p n ≥r$ if and only if

$pg ( a ) + ( 1 - p ) g ( b ) ≥g ( r )$
(27)

for all$a,b∈I$such that arb and apb1-p= r.

WLCF-Corollary. Let g be a function defined on a positive interval $I$ such that f(u) = g(eu ) is convex for $e u ≤r∈I$, aand let p1, p2, ..., p n be positive real numbers such that

$p=min { p 1 , p 2 , … , p n } , p 1 + p 2 +⋯+ p n =1.$
(28)

The inequality

$p 1 g ( a 1 ) + p 2 g ( a 2 ) +⋯+ p n g ( a n ) ≥g ( a 1 p 1 a 2 p 2 … a n p n )$
(29)

holds for all $a 1 , a 2 ,…, a n ∈I$ satisfying $a 1 p 1 a 2 p 2 … a n p n ≤r$ if and only if

$pg ( a ) + ( 1 - p ) g ( b ) ≥g ( r )$
(30)

for all$a,b∈I$such that arb and apb1-p= r.

WHCF-Corollary. Let g be a function defined on a positive interval $I$ such that f(u) = g(eu ) is convex for eur or eur, where $r∈I$, and let p1, p2, ..., p n be positive real numbers such that

$p=min { p 1 , p 2 , … , p n } , p 1 + p 2 +⋯+ p n =1.$
(31)

The inequality

$p 1 g ( a 1 ) + p 2 g ( a 2 ) +⋯+ p n g ( a n ) ≥g ( a 1 p 1 a 2 p 2 … a n p n )$
(32)

holds for all $a 1 , a 2 ,…, a n ∈I$ satisfying $a 1 p 1 a 2 p 2 … a n p n =r$ if and only if

$pg ( a ) + ( 1 - p ) g ( b ) ≥g ( r )$
(33)

for all$a,b∈I$such that apb1-p= r.

Remark 2.2. Let us denote

$g ( u ) = f ( u ) - f ( s ) u - s ,h ( x , y ) = g ( x ) - g ( y ) x - y .$
(34)

In some applications, it is useful to replace the hypothesis

$pf ( x ) + ( 1 - p ) f ( y ) ≥f ( s )$
(35)

in WRCF-, WLCF-, and WHCF-Theorems by the equivalent condition:

h(x, y) ≥ 0 for all $x,y∈I$ such that px + (1 - p)y = s.

This equivalence is true since

$p f ( x ) + ( 1 - p ) f ( y ) - f ( s ) = p [ f ( x ) - f ( s ) ] + ( 1 - p ) [ f ( y ) - f ( s ) ] = p ( x - s ) g ( x ) + ( 1 - p ) ( y - s ) g ( y ) = p ( 1 - p ) ( x - y ) [ g ( x ) - g ( y ) ] = p ( 1 - p ) ( x - y ) 2 h ( x , y ) .$

Remark 2.3. The required inequalities in WRCF-, WLCF-, and WHCF-Theorems turn into equalities for x1 = x2 = ... = x n . In addition, on the assumption that p1 = min{p1, p2, ..., p n }, equality holds for x1 = x and x2 = ... = x n = y if there exist $x,y∈I$, xy such that px + (1 - p)y = s and pf(x) + (1 - p)f(y) = f(s).

## 3 Proof of Lemma 2.1

Consider only the nontrivial case a < b. Since x1, x2, ..., x m [a, b] there exist λ1, λ2, ..., λ m [0, 1] such that

$x i = λ i a+ ( 1 - λ i ) b,i=1,2,…,m.$

From

$λ i = x i - b a - b ,i=1,2,…,m,$

we have

$∑ i = 1 m r i λ i = 1 a - b ( ∑ i = 1 m r i x i - b ∑ i = 1 m r i ) = 1 a - b [ q 1 a + q 2 b - b ( q 1 + q 2 ) ] = q 1 .$

Thus, according to Jensen's inequality, we get

$∑ i = 1 m r i f ( x i ) ≤ ∑ i = 1 m r i [ λ i f ( a ) + ( 1 - λ i ) f ( b ) ] = [ f ( a ) - f ( b ) ] ∑ i = 1 m r i λ i + f ( b ) ∑ i = 1 m r i = [ f ( a ) - f ( b ) ] q 1 + f ( b ) ( q 1 + q 2 ) = q 1 f ( a ) + q 2 f ( b ) .$

## 4 Proof of WRCF-Theorem

Since the necessity is obvious, we prove further the sufficiency. Without loss of generality, assume that x1x2 ≤ ... ≤ x n . If x1s, then the required inequality follows by Jensen's inequality for convex functions. Otherwise, since

$p 1 x 1 + p 2 x 2 + ⋯ + p n x n ≥ ( p 1 + p 2 + ⋯ + p n ) s ,$

there exists k {1, 2, ..., n - 1} such that

$x 1 ≤⋯≤ x k

Let us denote

$q = p 1 + ⋯ + p k ,$

By Jensen's inequality, we have

$∑ i = k + 1 n p i f ( x i ) )≥ ( p k + 1 + ⋯ + p n ) f ( z ) = ( 1 - q ) f ( z ) ,$

where

$z= p k + 1 x k + 1 + ⋯ + p n x n p k + 1 + ⋯ + p n ,z≥s,z∈I.$

Thus, it suffices to prove that

$∑ i = 1 k p i f ( x i ) + ( 1 - q ) f ( z ) ≥ f ( S ) ,$
(36)

where

$S= p 1 x 1 + p 2 x 2 +⋯+ p n x n = p 1 x 1 + p 2 x 2 + ⋯ + p n x n p 1 + p 2 + ⋯ + p n ,s≤S≤z.$

Let y i , i = 1, 2, ..., k, defined by

$p x i + ( 1 - p ) y i =s.$

We will show that

$z ≥ y 1 ≥ y 2 ≥ ⋯ ≥ y k > s .$

We have

$y 1 ≥ y 2 ≥ ⋯ ≥ y k , y k - s = p ( s - x k ) 1 - p > 0 , y 1 = s - p x 1 1 - p ≤ S - p x 1 1 - p = ( p 1 - p ) x 1 + p 2 x 2 + ⋯ + p n x n ( p 1 - p ) + p 2 + ⋯ + p n .$

Since p1 - p = p1 - min {p1, p2, ..., p n } ≥ 0, we get

$( p 1 - p ) x 1 + p 2 x 2 + ⋯ + p n x n ( p 1 - p ) + p 2 + ⋯ + p n ≤ p 2 x 2 + ⋯ + p n x n p 2 + ⋯ + p n ≤z,$

and hence y1z. Now, from zy1y2 ≥ ... ≥ yk > s, it follows that $y 1 , y 2 ,…, y k ∈I$. Then, by hypothesis, we have

$p f ( x i ) + ( 1 - p ) f ( y i ) ≥ f ( s )$

for i = 1,2, ..., k. Summing all these inequalities multiplied by p i /p, respectively, we get

$∑ i = 1 k p i f ( x i ) + 1 - p p ∑ i = 1 k p i f ( y i ) ≥ q p f ( s ) .$

Therefore, to prove (36), it suffices to show that

$q p f ( s ) + ( 1 - q ) f ( z ) ≥ 1 - p p ∑ i = 1 k p i f ( y i ) +f ( S ) .$
(37)

Since S [s, z], y i (s, z] for i = 1, 2, ..., k,

$q p + 1 - q = 1 - p p ∑ i = 1 k p i + 1 , q p s + ( 1 - q ) z = 1 - p p ∑ i = 1 k p i y i + S ,$
1. (37)

is a consequence of Lemma 2.1, where m = k + 1, a = s, b = z, q 1 = q/p, q 2 = 1 - q, r m = 1, x m = S, r i = (1 - p)p i /p and x i = y i for i = 1, 2, ..., k.

## 5 Applications

Proposition 5.1. Let a1, a2, ..., a n (n ≥ 3) be positive real numbers such that a1a2 ... a n = 1. If p and q are nonnegative real numbers such that p + qn - 1, then

$1 1 + p a 1 + q a 1 2 + 1 1 + p a 2 + q a 2 2 + ⋯ + 1 1 + p a n + q a n 2 ≥ n 1 + p + q .$

Proof. Write the desired inequality as

$g ( a 1 ) +g ( a 2 ) +⋯+g ( a n ) ≥ng ( 1 ) ,$

where

$g ( t ) = 1 1 + p t + q t 2 ,t>0.$

To prove this inequality, we apply HCF-Corollary for r = 1. Let

$f ( u ) =g ( e u ) = 1 1 + p e u + q e 2 u ,u∈ℝ.$

Using the second derivative,

$f ″ ( u ) = e u [ 4 q 2 e 3 u + 3 p q e 2 u + ( p 2 - 4 q ) e u - p ] ( 1 + p e u + q e 2 u ) 3 ,$

we will show that f(u) is convex for eur = 1. We need to show that

$4 q 2 t 3 +3pq t 2 + ( p 2 - 4 q ) t-p≥0$

for t ≥ 1. Indeed,

$4 q 2 t 3 + 3 p q t 2 + ( p 2 - 4 q ) t - p ≥ ( 4 q 2 + 3 p q + p 2 - 4 q - p ) t = [ ( p + 2 q ) ( p + q - 2 ) + 2 q 2 + p ] t > 0 ,$

because p + qn - 1 ≥ 2.

By HCF-Corollary, it suffices to prove that g(a) + (n - 1)g(b) ≥ ng(1) for all a, b > 0 such that abn-1= 1. We write this inequality as

$b 2 n - 2 b 2 n - 2 + p b n - 1 + q + n - 1 1 + p b + q b 2 ≥ n 1 + p + q .$

Applying the Cauchy-Schwarz inequality, it suffices to prove that

$( b n - 1 + n - 1 ) 2 ( b 2 n - 2 + p b n - 1 + q ) + ( n - 1 ) ( 1 + p b + q b 2 ) ≥ n 1 + p + q ,$

which is equivalent to

$pB+qC≥A,$

where

$A = ( n - 1 ) ( b n - 1 - 1 ) 2 ≥ 0 , B = ( b n - 1 - 1 ) 2 + n E = A n - 1 + n E , C = ( b n - 1 - 1 ) 2 + n F = A n - 1 + n F ,$

with

$E= b n - 1 +n-2- ( n - 1 ) b,F=2 b n - 1 +n-3- ( n - 1 ) b 2 .$

By the AM-GM inequality applied to n - 1 positive numbers, we have E ≥ 0 and F ≥ 0 for n ≥ 3. Since A ≥ 0 and p + qn - 1, we have

$pB+qC-A≥pB+qC- ( p + q ) A n - 1 =n ( p E + q F ) ≥0.$

Equality holds for a1 = a2 = ... = a n = 1.

Remark 5.2. For p + q = n - 1 and n ≥ 3, by Proposition 5.1 we get the following beautiful inequality

$1 1 + p a 1 + q a 1 2 + 1 1 + p a 2 + q a 2 2 +⋯+ 1 1 + p a n + q a n 2 ≥1.$

If p = n - 1 and q = 0, then we get the well-known inequality

$1 1 + ( n - 1 ) a 1 + 1 1 + ( n - 1 ) a 2 +⋯+ 1 1 + ( n - 1 ) a n ≥1.$

Remark 5.3. For $p=q= 1 r$, $0 and n ≥ 3, by Proposition 5.1 we obtain the inequality

$1 r + a 1 + a 1 2 + 1 r + a 2 + a 2 2 + ⋯ + 1 r + a n + a n 2 ≥ n r + 2 .$

In addition, for $r= 2 n - 1$, n ≥ 3, we get

$∑ i = 1 n 1 2 + ( n - 1 ) ( a i + a i 2 ) ≥ 1 2 .$

Remark 5.4. For p = 2r, q = r2, $r≥ n -1$ and n ≥ 3, by Proposition 5.1 we obtain

$1 ( 1 + r a 1 ) 2 + 1 ( 1 + r a 2 ) 2 +⋯+ 1 ( 1 + r a n ) 2 ≥ n ( 1 + r ) 2 .$

Proposition 5.5. Let a1, a2, ..., a n (n ≥ 4) be positive real numbers such that a1a2 ... a n = 1. If p, q, r are nonnegative real numbers such that p + q + rn - 1, then

$∑ i = 1 n 1 1 + p a i + q a i 2 + r a i 3 ≥ n 1 + p + q + r .$

Proof. Write the required inequality as

$g ( a 1 ) + g ( a 2 ) + ⋯ + g ( a n ) ≥ n g ( 1 ) ,$

where

$g ( t ) = 1 1 + p t + q t 2 + r t 3 , t > 0 ,$

and apply HCF-Corollary to g(t) for r = 1. Let

$f ( u ) = g ( e u ) = 1 1 + p e u + q e 2 u + r e 3 u ,$

defined on . For n ≥ 4, which implies p + q + r ≥ 3, we claim that f is convex for eu ≥ 1. Since

$f ″ ( u ) = t [ 9 r 2 t 5 + 1 1 q r t 4 + ( 2 p r + 4 q 2 ) t 3 + ( 3 p q - 9 r ) t 2 + ( p 2 - 4 q ) t - p ] ( 1 + p t + q t 2 + r t 3 ) 3 ,$

where t = eu ≥ 1, we need to show that

$9 r 2 t 5 + 1 1 q r t 4 + ( 2 p r + 4 q 2 ) t 3 + ( 3 p q - 9 r ) t 2 + ( p 2 - 4 q ) t - p ≥ 0$

Since

$9 r 2 t 5 + 1 1 q r t 4 + ( 2 p r + 4 q 2 ) t 3 - p ≥ ( 9 r 2 + 1 1 q r + 2 p r + 4 q 2 ) t 3 - p t ,$

it suffices to show that

$( 9 r 2 + 1 1 q r + 2 p r + 4 q 2 ) t 2 + ( 3 p q - 9 r ) t + p 2 - p - 4 q ≥ 0 .$

Using the inequality t2 ≥ 2t - 1, we still have to prove that At + B ≥ 0, where

$A = 1 8 r 2 + 2 2 q r + 4 p r + 8 q 2 + 3 p q - 9 r , B = - 9 r 2 - 1 1 q r - 2 p r - 4 q 2 + p 2 - p - 4 q .$

Since p + q + r ≥ 3, we have

$A ≥ 1 8 r 2 + 2 2 q r + 4 p r + 8 q 2 + 3 p q - 3 r ( p + q + r ) = 1 5 r 2 + 1 9 q r + p r + 8 q 2 + 3 p q ≥ 0 .$

Therefore,

$A t + B ≥ A + B = p 2 + 4 q 2 + 9 r 2 + 3 p q + 1 1 q r + 2 p r - ( p + 4 q + 9 r ) ≥ p 2 + 4 q 2 + 9 r 2 + 3 p q + 1 1 q r + 2 p r - ( p + 4 q + 9 r ) ( p + q + r ) 3 = 2 ( p - r ) 2 + 9 q 2 + 1 6 r 2 + 4 p q + 2 0 q r 3 > 0 .$

According to HCF-Corollary, it suffices to prove that g(a) + (n - 1) g(b) ≥ ng(1) for all a, b > 0 such that abn-1= 1. We write this inequality as

$b 3 n - 3 b 3 n - 3 + p b 2 n - 2 + q b n - 1 + r + n - 1 1 + p b + q b 2 + r b 3 ≥ n 1 + p + q + r ,$

or

$p 2 A 1 1 + q 2 A 2 2 + r 2 A 3 3 + p q A 1 2 + q r A 2 3 + r p A 3 1 ≥ A p + B q + C r ,$

where

$A 1 1 = b 2 n - 2 ( b n - n b + n - 1 ) , A 2 2 = b n - 1 ( b 2 n - n b 2 + n - 1 ) , A 3 3 = b 3 n - n b 3 + n - 1 ,$
$A 1 2 = b 3 n - 1 + b 3 n - 2 + ( n - 1 ) ( b 2 n - 2 + b n - 1 ) - n ( b 2 n + b n ) , A 2 3 = b 3 n + b 3 n - 1 + ( n - 1 ) ( b n - 1 + 1 ) - n ( b n + 2 + b 2 ) , A 3 1 = b 3 n + b 3 n - 2 + ( n - 1 ) ( b 2 n - 2 + 1 ) - n ( b 2 n + 1 + b ) ,$
$A = b 2 n - 2 [ ( n - 1 ) b n - n b n - 1 + 1 ] , B = b n - 1 [ ( n - 1 ) b 2 n - n b 2 n - 2 + 1 ] , C = ( n - 1 ) b 3 n - n b 3 n - 3 + 1 .$

Since A,B,C ≥ 0 (from the AM-GM inequality applied to n positive numbers) and p + q + rn - 1, it suffices to show that

$( n - 1 ) ( p 2 A 1 1 + q 2 A 2 2 + r 2 A 3 3 + p q A 1 2 + q r A 2 3 + r p A 3 1 ) ≥$

which is equivalent to

$p 2 B 1 1 + q 2 B 2 2 + r 2 B 3 3 + p q B 1 2 + q r B 2 3 + r p B 3 1 ≥ 0 ,$
(38)

where

$B 1 1 = ( n - 1 ) A 1 1 - A = n b 2 n - 2 [ b n - 1 - ( n - 1 ) b + n - 2 ] , B 2 2 = ( n - 1 ) A 2 2 - B = n b n - 1 [ b 2 n - 2 - ( n - 1 ) b 2 + n - 2 ] , B 3 3 = ( n - 1 ) A 3 3 - C = n [ b 3 n - 3 - ( n - 1 ) b 3 + n - 2 ] ,$
$B 1 2 = ( n - 1 ) A 1 2 - A - B = n b n - 1 [ 2 b 2 n - 2 - ( n - 1 ) b n + 1 + ( n - 2 ) b n - 1 - ( n - 1 ) b + n - 2 ] = n b 2 n - 2 [ 2 b n - 1 - ( n - 1 ) b 2 + n - 3 ] + n b n - 1 [ b n - 1 - ( n - 1 ) b + n - 2 ] ,$
$B 2 3 = ( n - 1 ) A 2 3 - B - C = n [ 2 b 3 n - 3 - ( n - 1 ) b n + 2 + ( n - 2 ) b n - 1 - ( n - 1 ) b 2 + n - 2 ] ,$
$B 3 1 = ( n - 1 ) A 3 1 - C - A = n [ 2 b 3 n - 3 - ( n - 1 ) b 2 n + 1 + ( n - 2 ) b 2 n - 2 - ( n - 1 ) b + n - 2 ] .$

We see that B11, B22, B33, B12 ≥ 0 (by the AM-GM inequality applied to n - 1 positive numbers). Also, we have

$B 2 3 n = b n - 1 [ 3 b n - 1 - ( n - 1 ) b 3 + n - 4 ] + 2 b n - 1 ( b n - 1 - 1 ) 2 + b 2 n - 2 - ( n - 1 ) b 2 + n - 2 ≥ 0 ,$

since

$3 b n - 1 - ( n - 1 ) b 3 +n-4≥0, b 2 n - 2 - ( n - 1 ) b 2 +n-2≥0$

(by the AM-GM inequality applied to n - 1 positive numbers). Using the inequality bn-1- (n - 1)b + n - 2 ≥ 0,we get B31D, where

$D=n b n - 1 [ 2 b 2 n - 2 - ( n - 1 ) b n + 2 + ( n - 2 ) b n - 1 - 1 ] .$

To prove (38), it suffices to show that p2B11 + r2B33 + prD ≥ 0. This is true if 4B11B33D2; that is,

$4 [ b n − 1 − ( n − 1 ) b + n − 2 ] [ b 3 n − 3 − ( n − 1 ) b 3 + n − 2 ] ≥ ≥ [ 2 b 2 n − 2 − ( n − 1 ) b n + 2 + ( n − 2 ) b n − 1 − 1 ] 2 .$
(39)

In the case n = 4, (39) becomes in succession

$4 ( b 3 − 3 b + 2 ) ( b 9 − 3 b 3 + 2 ) ≥ ( b 6 − 2 b 3 + 1 ) 2 , 4 ( b − 1 ) 2 ( b + 2 ) ( b 3 − 1 ) 2 ( b 3 + 2 ) ≥ ( b 3 − 1 ) 4 , ( b − 1 ) 2 ( b 3 − 1 ) 2 ( 3 b 4 + 5 b 3 − 3 b 2 + 6 b + 15 ) ≥ 0 .$

Clearly, the last inequality is true. The inequality (39) also holds for n ≥ 5, but we leave this to the reader to prove. Equality occurs for a1 = a2 = ... = a n = 1.

Remark 5.6. For n = 4 and p + q + r = 3, by Proposition 5.5 we get the following beautiful inequality

$∑ i = 1 4 1 1 + p a i + q a i 2 + r a i 3 ≥1.$

In addition, for p = q = r = 1, we get the known inequality ()

$∑ i = 1 4 1 1 + a i + a i 2 + a i 3 ≥1.$

Conjecture 5.7. Let a1,a2, ..., a n be positive real numbers such that a1a2 ... a n = 1, and let k1,k2, ..., k m be nonnegative real numbers such that k1 + k2 + ... + k m n - 1. If mn - 1, then

$∑ i = 1 n 1 1 + k 1 a i + k 2 a i 2 + ⋯ + k m a i m ≥ n 1 + k 1 + k 2 + ⋯ + k m .$
(40)

Remark 5.8. For m = n - 1 and k1 = k2 = ... = k m = 1, (40) turns into the known beautiful inequality ()

$∑ i = 1 n 1 1 + a i + a i 2 + ⋯ + a i n - 1 ≥1.$

Remark 5.9. For $k 1 = ( m 1 ) r$,$k 2 = ( m 2 ) r 2 , … , k m = ( m m ) r m$, (40) turns into the known inequality [1, 2]

$∑ i = 1 n 1 ( 1 + r a i ) m ≥ n ( 1 + r ) m ,$

which holds for 1 ≤ mn - 1 and $r≥ n m -1$. □

Proposition 5.10. If x1, x2, ..., x n are nonnegative real numbers such that

$x 1 +2 x 2 +⋯+n x n = n ( n + 1 ) 2 ,$

then

$( n - 1 ) ( n + 2 ) x 1 3 + 2 x 2 3 + ⋯ + n x n 3 - n ( n + 1 ) 2 ≥ ≥ 2 ( n 2 + n - 1 ) x 1 2 + 2 x 2 2 + ⋯ + n x n 2 - n ( n + 1 ) 2 .$

Proof. Since the inequality is trivial for n = 1, consider further that n ≥ 2. Write the inequality as

$p 1 f ( x 1 ) + p 2 f ( x 2 ) +⋯+ p n f ( x n ) ≥f ( p 1 x 1 + p 2 x 2 + ⋯ + p n x n ) ,$

where

$p i = 2 i n ( n + 1 ) , i = 1 , 2 , … , n , f ( u ) = ( n - 1 ) ( n + 2 ) u 3 - 2 ( n 2 + n - 1 ) u 2 , u ≥ 0 .$

The function f(u) is convex for us = 1, since

$f ″ ( u ) = 6 ( n - 1 ) ( n + 2 ) u - 4 ( n 2 + n - 1 ) ≥ 6 ( n - 1 ) ( n + 2 ) - 4 ( n 2 + n - 1 ) = 2 ( n 2 + n - 4 ) > 0$

for u ≥ 1. According to WHCF-Theorem and Remark 2.2, it suffices to prove that h(x, y) ≥ 0 for all x, y ≥ 0 such that px + (1 - p)y = 1, where

$p=min { p 1 , p 2 , … , p n } = 2 n ( n + 1 ) .$

We have

$g ( u ) = f ( u ) - f ( 1 ) u - 1 = ( n - 1 ) ( n + 2 ) ( u 2 + u + 1 ) - 2 ( n 2 + n - 1 ) ( u + 1 ) , h ( x , y ) = g ( x ) - g ( y ) x - y = ( n - 1 ) ( n + 2 ) ( x + y + 1 ) - 2 ( n 2 + n - 1 ) .$

From px + (1 - p)y = 1, we get

$x + y = x + 1 - p x 1 - p = 1 + ( 1 - 2 p ) x 1 - p = n ( n + 1 ) + ( n 2 + n - 4 ) x ( n - 1 ) ( n + 2 ) ≥ n ( n + 1 ) ( n - 1 ) ( n + 2 ) ,$

and hence

$h ( x , y ) ≥ ( n - 1 ) ( n + 2 ) n ( n + 1 ) ( n - 1 ) ( n + 2 ) + 1 -2 ( n 2 + n - 1 ) =0.$

This completes the proof. Equality holds for a1 = a2 = ... a n = 1, and also for a1 = 0 and $a 2 =⋯= a n = n ( n + 1 ) ( n - 1 ) ( n + 2 ) .$

## References

1. 1.

Cirtoaje V: A generalization of Jensen's inequality. Gazeta Matematica Seria A 2005, 2: 124–138.

2. 2.

Cirtoaje V: Algebraic Inequalities--Old and New Methods. GIL Publishing House, Zalau 2006.

3. 3.

Tetiva M: A new proof for the right convex function theorem. Gazeta Matematica Seria A 2006, 2: 126–133.

4. 4.

Art of Problem Solving[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=223266]

## Acknowledgements

The authors are grateful to the referees for their useful comments.

## Author information

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### Corresponding author

Correspondence to Vasile Cirtoaje.

### Competing interests

The authors declare that they have no competing interests.

### Authors' contributions

VC conceived and proved the main results and their applications. AB performed numerical verification for all applications and prepared the first manuscript. Both authors read and approved the final manuscript.

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Cirtoaje, V., Baiesu, A. An extension of Jensen's discrete inequality to half convex functions. J Inequal Appl 2011, 101 (2011). https://doi.org/10.1186/1029-242X-2011-101 