it's related to my OP here A specific Jensen's inequality for a proof

We have for $a+b+c=3$ where $a,b,c$ are positive real numbers

$ \sqrt{2} \leq \displaystyle\sum_{cyc} \frac{\sqrt{a^2+b^2}}{2ab+1}$

Michael Rozenberg have proved this with elegance .

So I was wondering if there exists a generalization to this (with the same condition) and we have :

$$ \frac{3\sqrt{\alpha}\sqrt{2}}{2\beta +1} \leq \displaystyle\sum_{{a,b,c}} \frac{\sqrt{\alpha(a^2+b^2)}}{2\beta ab+1}$$

The sum above is cyclic.$\alpha,\beta$ are positive real numbers .

I have no idea to prove this and I get this with my method related to my previous OP .

Thanks a lot .

  • $\begingroup$ I have a proof for all $\beta\geq\frac{1}{18}$ and $\beta=0$. $\endgroup$ – Michael Rozenberg Jan 1 '18 at 15:32
  • $\begingroup$ @Michael Rozenberg can you tell me if the minimum of :$$\frac{(2ab+1)}{\sqrt{a^2+b^2}}+ \frac{(2cb+1)}{\sqrt{c^2+b^2}}+ \frac{(2ac+1)}{\sqrt{a^2+c^2}}$$ is easier to calculate ? If yes I have a full proof .Thanks a lot. $\endgroup$ – user448747 Jan 2 '18 at 20:34
  • $\begingroup$ For positives $a$, $b$ and $c$ such that $a+b+c=3$? $\endgroup$ – Michael Rozenberg Jan 2 '18 at 20:37
  • $\begingroup$ Yes sorry I forgot the condition wich is the same . $\endgroup$ – user448747 Jan 2 '18 at 20:38
  • $\begingroup$ I see that it's not $\frac{9}{\sqrt2}.$ $\endgroup$ – Michael Rozenberg Jan 2 '18 at 20:42

Your Answer

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