New bounds for convex function of 2 variables

It's related to my post : New bound for Am-Gm of 2 variables

In fact I have discovered (maybe it's already knew) a new formula for convex function this is the following :

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

1) $$f(x)\geq 0\quad \forall x \in I$$

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

3)$$f(x)\neq \text{constant function}$$

Then we have for $$x,y \in I$$: $$\sqrt{f\Big(\frac{xf'(x)+yf'(y)}{f'(x)+f'(y)}\Big)f\Big(\frac{x+y}{2}\Big)}\leq \frac{f(x)+f(y)}{2}\leq \frac{f\Big(\frac{xf'(x)+yf'(y)}{f'(x)+f'(y)}\Big)+f\Big(\frac{x+y}{2}\Big)}{2}$$

For the RHS we can use Karamata's inequality , the first line of the majorization is easy to check the second line gives :

$$x+y\leq \frac{xf'(x)+yf'(y)}{f'(x)+f'(y)}+\frac{x+y}{2}$$

Or after manipulation:

$$\frac{x+y}{2}(f'(x)+f'(y))\leq xf'(x)+yf'(y)$$

Wich is true because $$f'(x)$$ is increasing ($$f''(x)\geq 0$$) .

The LHS is more delicate and it's the reason why I post here . Maybe we can introduce the logarithm and see what happend...

So if you have any hints it would be nice (or a counter-example) .

Thanks in advance for your time .

Ps: For the story I'm inspired by Slater's inequality and Jensen's inequality .

• Once more I am confused by your formulation: You have “discovered a new formula” and then ask for “hints or a counter-example”. So you are guessing the formula, or what? – Martin R Jun 13 at 13:16

The left hand side of the inequality fails for $$f(x)=e^{x^2}$$, $$x=0$$, $$y=1$$ , and $$I=[x,y]$$. Indeed, $$f’(t)=2te^{t^2}$$ for each real $$t$$. So $$f’(x)=0$$ and the left hand side of the inequality transforms to a false inequality $$\sqrt{f(1)f\left(\frac 12\right)}\le \frac {1+f(1)}2$$ or $$\sqrt{e\cdot e^{1/4}}=1.868\dots \le 1.859\dots=\frac {1+e}2$$.

The inequality also fails for $$x=\tfrac 12$$ and $$y=\tfrac 34$$, see the following Mathcad calculations. • I like to work with you but just something does it works if we add the condition $f'(x)\geq f(x)\forall x \in I$? – user674646 Jun 22 at 12:38
• @FatsWallers With this condition the inequality also can fail, see the updated answer. – Alex Ravsky Jun 22 at 14:26
• Okay many thanks ! – user674646 Jun 22 at 14:56
• have you found a counter-example for the RHS ? – user674646 Jun 29 at 17:00
• @FatsWallers As I understood from the question, you have proved RHS, no? – Alex Ravsky Jun 29 at 17:27

In a comment to another question by FatsWallers, Paata Ivanishvili pointed out that the right inequality is wrong too. In fact, it is false for $$f(x)=x^3+ax$$ for every positive $$a$$ and $$I=[0,\sqrt{a}]$$. Setting $$x=\sqrt{a}/3$$, $$y=2\sqrt{a}/3$$, the RHS minus the LHS is $$-19/31944\cdot\sqrt{a^3}<0$$.