# A difficult generalization of an inequality

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 .

• I have a proof for all $\beta\geq\frac{1}{18}$ and $\beta=0$. – Michael Rozenberg Jan 1 '18 at 15:32
• @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. – user448747 Jan 2 '18 at 20:34
• For positives $a$, $b$ and $c$ such that $a+b+c=3$? – Michael Rozenberg Jan 2 '18 at 20:37
• Yes sorry I forgot the condition wich is the same . – user448747 Jan 2 '18 at 20:38
• I see that it's not $\frac{9}{\sqrt2}.$ – Michael Rozenberg Jan 2 '18 at 20:42