We have the following conjecture :

Let $x_i$ be $n$ real numbers and $p_i>0$ be $n$ weights and $f(x)=e^{x}$ then we have : $$\Big(\sum_{i=1}^{n}p_i\Big)\Big(f\Big(\frac{\sum_{i=1}^{n}x_ip_i}{\sum_{i=1}^{n}p_i}\Big)f\Big(\frac{\sum_{i=1}^{n}x_if(x_i)p_i}{\sum_{i=1}^{n}p_if(x_i)}\Big)\Big)^{\frac{1}{2}}\leq \sum_{i=1}^{n}p_if(x_i)$$

This inequality is more precise than the classical Jensen's inequality with exponential .

My try :

I think we can use induction to solve this problem for $n=2$ we have :

$$\Big(p_1+p_2\Big)\Big(f\Big(\frac{x_1p_1+x_2p_2}{p_1+p_2}\Big)f\Big(\frac{x_1f(x_1)p_1+x_2f(x_2)p_2}{p_1f(x_1)+p_2f(x_2)}\Big)\Big)^{\frac{1}{2}}\leq p_1f(x_1)+p_2f(x_2)$$

But after this I'm really stuck...

Thanks in advance for your time.

  • 2
    $\begingroup$ Do you have reason to believe this conjecture is actually true? $\endgroup$ – David M. Mar 8 at 13:28
  • $\begingroup$ @David M. My argument is : Slater's inequality is more precise than Jensen's inequality if we consider just the exponential function so the geometric mean of the two quantities is inferior to the sum of exponential with weight ... $\endgroup$ – user635269 Mar 8 at 13:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy