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I'm confusing because in the inequality New bounds for convex function of 2 variables the RHS doesn't work (take $f(x)=\tan(x)$) so I propose a new formula that I have checked for all the elementary convex functions . I recognize that I'm inspired like here by the upper estimate of Jensen's inequality found by C.p Niculescu .To recall we have also the Slater's inequality wich is the compagnion of the Jensen's inequality . I used it to give a new estimate like this .

Let $f(x)$ be a twice differentiable function on an interval $I$ such that :

$1)$ $f'(x)>0$ $\forall x \in I $

$2)$ $f''(x)>0$ $\forall x \in I $

Then for $a,b,c,d\in I$ such that $a\geq b\geq c\geq d $ we have : $$|f(a)+f(d)-f\Big(\frac{af'(a)+df'(d)}{f'(a)+f'(d)}\Big)-f\Big(\frac{a+d}{2}\Big)|\geq |f(b)+f(c)-f\Big(\frac{bf'(b)+cf'(c)}{f'(b)+f'(c)}\Big)-f\Big(\frac{b+c}{2}\Big)|$$

Furthermore I add absolute value because the inequality can be reversed as we know .

My question is : Have you a counter-example ?

Many thanks .

Community
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  • In your another post, you use Karamata's inequality to obtain that here the stuff in the absolute sign is non-positive, right? – River Li Jul 13 '19 at 04:51

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Given $f$, I used Mathcad to draw a (two-dimensional) graph of a function $$h(x,y)=\left|f(x)+f(y)-f\Big(\frac{xf'(x)+yf'(y)}{f'(x)+f'(y)}\Big)-f\Big(\frac{x+y}{2}\Big)\right|$$ and then looked for “monotonicity violation” of $h$. The second try was $f(x)=x^3$ and I quickly found that $$h(0.2,10)\simeq 131.468>129.168\simeq h(0.01,10.1).$$

Alex Ravsky
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