Prove that: $\sum\limits_{cyc}\frac{1}{\sqrt{2a^2+5ab+2b^2}} \geq\sqrt{\frac{3}{ab+ac+bc}}$ Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$ \dfrac{1}{\sqrt{2a^2+5ab+2b^2}}+\dfrac{1}{\sqrt{2b^2+5bc+2c^2}}+\dfrac{1}{\sqrt{2c^2+5ca+2a^2}} \geq\sqrt{\frac{3}{ab+ac+bc}}.$$
I solved this problem by Hölder:
$$\left(\sum_{cyc}\dfrac{1}{\sqrt{2a^2+5ab+2b^2}}\right)^2\sum_{cyc}\frac{(a+b)^3}{(2a^2+5ab+2b^2)^2}\geq\left(\sum_{cyc}\frac{a+b}{2a^2+5ab+2b^2}\right)^3$$
and it remains to prove that
$$(ab+ac+bc)\left(\sum_{cyc}\frac{a+b}{2a^2+5ab+2b^2}\right)^3\geq3\sum_{cyc}\frac{(a+b)^3}{(2a^2+5ab+2b^2)^2},$$ which is true by BW and by using computer.
In this topic https://artofproblemsolving.com/community/c6h542992 there is a proof (from gxggs), but it's very very complicated.
I found another way, a smooth enough, but it's still a very hard solution.
I am looking for a nice proof by hand, for which there is a possibility to find this proof during a competition.
Thank you!

My attempt for this very hard inequality as conjecture   :
Conjecture :
Let $x\geq z$ and $z\leq y\leq 1.5z$ and $z\geq 1$ then it seems we have :
$$0\le f\left(x\right)+f\left(y\right)-2f\left(\frac{2}{\frac{1}{x}+\frac{1}{y}}\right)$$
Where : $$f\left(a\right)=\frac{1}{\sqrt{2a^{2}+z\left(a-z\right)}}$$
The trick here is to use $a+b=u$ and $ab=b(u-b)$
Next it seems easier to show for $x,b,c>0$:
$$g\left(x\right)=\frac{2}{\sqrt{2\left(\frac{2}{\frac{1}{x+b}+\frac{1}{b+c}}\right)^{2}+b\left(\frac{2}{\frac{1}{x+b}+\frac{1}{b+c}}-b\right)}}+\frac{1}{\sqrt{2x^{2}+2c^{2}+5xc}}-\sqrt{\frac{3}{xb+bc+cx}}\geq 0$$
Hope it helps !
