<s>[INST] A specific Jensen's inequality for a proof It's related to this Inequality : $\sum_{cyc}\frac{\sqrt{2}a^2b}{2a+b} \leq \sum_{cyc} \frac{\sqrt{a^2+b^2}}{2ab+1}$ more particulary to this passage :
Please suggest how to show that
$ \sqrt{2} \leq \displaystyle\sum_{cyc} \frac{\sqrt{a^2+b^2}}{2ab+1}$
My try
There is a simple method ,you can minimize each term like this but first you have:
$$\frac{\sqrt{a^2+b^2}}{2ab+1}$$
We can rewrite the expression like this :
$$\frac{\sqrt{(a+b)^2-2ab}}{2ab+1}$$
Now the idea is to remark that for a fixed sum of two numbers the product of this two numbers is maximal when each term is equal .
Here we use this method and remark  that this method minimize the numerator and maximize the denominator so we have :
$a+b=x$ and $ab=0.25x^2$
$c+b=y$ and $bc=0.25y^2$
$a+c=z$ and $ac=0.25z^2$
We get :
$$\frac{\sqrt{x^2-0.5x^2}}{0.5x^2+1}+\frac{\sqrt{y^2-0.5y^2}}{0.5y^2+1}+\frac{\sqrt{z^2-0.5z^2}}{0.5z^2+1}$$
So we have to prove :
$$\frac{\sqrt{x^2-0.5x^2}}{0.5x^2+1}+\frac{\sqrt{y^2-0.5y^2}}{0.5y^2+1}+\frac{\sqrt{z^2-0.5z^2}}{0.5z^2+1}\geq \sqrt{2}$$
But I can't use directly Jensen because the function $f(x)=\frac{\sqrt{x^2-0.5x^2}}{0.5x^2+1}$ is not convex so my question is :

1)What's the kind of function $f(x)$ is ?
2)What's the specific Jensen's inequality we have to use ?

Thanks a lot .
 [/INST] I think your reasoning is total wrong.
It remains to prove that
$$\sum\frac{a+b}{\frac{(a+b)^2}{2}+1}\geq2,$$ which is wrong for $c=3$ and $a=b\rightarrow0^+.$
I have a proof of your starting inequality by full expanding, by BW, by uvw and by SOS (which you saw).
Which way do you prefer?
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