Prove or disprove the following generalization :$\sum_{i,j=1}^{n}f\left(a_{i}a_{j}\right)>n^{2}f\left(\sum_{i,j=1}^{n}\frac{a_{i}a_{j}}{n^{2}}\right)$

Question

It's a follow up generalization of An alternative lower bound for $\prod_{ i,j = 1}^n\frac{1+a_ia_j}{1-a_ia_j} $

Conjecture :

Let $f(x)$ be on $-1<x<1$ a strictly increasing function such that :

$1)f(x)=-f(-x)$

$2)f(1)=\infty$

$3)\forall x \in[0,1) f''(x)\geq 0,\forall x \in(-1,0] f''(x)\leq 0$

$4)\forall x\in (-1,1) f'''(x)>0$

Then for $a_i\in(-1,1)$ we have :

$$\sum_{i,j=1}^{n}f\left(a_{i}a_{j}\right)>n^{2}f\left(\sum_{i,j=1}^{n}\frac{a_{i}a_{j}}{n^{2}}\right)$$

As hint for a proof I used Fuch's inequality which is in itself an extension of Karamata's inequality for example in the case $n=2$ and $sgn(ab)=-1,1>a^2\geq b^2\geq ab>-1$ the vector $(a^2,b^2,-ab,-ab)$ majorize the vector $\left(\frac{a^{2}+b^{2}+2ab}{4},\frac{a^{2}+b^{2}+2ab}{4},-\frac{a^{2}+b^{2}+2ab}{4},-\frac{a^{2}+b^{2}+2ab}{4}\right)$ with weight $(1,1,-1,-1)$

For the general case remark that at some point the majorization decreases (decreasing minus increasing )and then at the end it's equal to zero .

Edit :

The draft is incorrect but we can use majorization in symmetric form see https://arxiv.org/abs/0803.2958

Is this draft correct ? If not have you a proof or a counter-example ?

Answer 1

Case $n=2$ :

If $a_ia_j\geq 0$ :

It's a simple application of Jensen's inequality .

If $a_ia_j\leq 0 $ :

Then we write the inequality as we have $f(0)=0,f(x)=-f(-x)$ :

$$f\left(a_{1}^{2}\right)+f\left(a_{2}^{2}\right)+4f\left(0\right)>4f\left(\frac{\left(a_{1}+a_{2}\right)^{2}}{4}\right)+2f\left(-a_{1}a_{2}\right)$$

Then we use the Darij's paper (linked in the question) with majorzation in symmetric form we have :

Let :

$$f\left(x,y\right)=\left|x^{2}-t\right|+\left|y^{2}-t\right|-4\left|\frac{x^{2}+y^{2}+2xy}{4}-t\right|-2\left|-xy-t\right|+4\left|t\right|$$

Then it's not hard to show that for $-1\leq a\leq 0\leq x\leq 1,t\in[0,1]$ we have :

$$f(x,a)\geq 0$$

Then apply Karamata's inequality .

We are done in the case $n=2$

Case $n=3$ and further

Restating the case $n=2$ if $a_ia_j\geq 0$ it's Jensen's inequality .

If $a_ia_j\leq 0$ then define :

$$f(x,y)=|x-y|,g(x)=f\left(a^{2},x\right)+f\left(b^{2},x\right)+f\left(c^{2},x\right)-9f\left(\frac{\left(a+b+c\right)^{2}}{9},x\right)+2f\left(ab,x\right)-2f\left(-bc,x\right)-2f\left(-ac,x\right)+8\left|x\right|$$

Then if $a,b\in(-1,0]$,such that $b=-c$ or $a=-c$ then :

$$g(x)\geq g(-x)=0, 0\leq x\leq 1$$

So the majorization in symmetric form is complete and then by Karamata's ineqality we have this special case . The general case in this specific form can be done as this way .

Then as the function is right convex we can apply the theorem here https://journalofinequalitiesandapplications.springeropen.com/articles/10.1186/1029-242X-2011-101

We are done .

Reference :

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