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!
 A: Remark: I give a proof using the so-called isolated fudging.
It suffices to prove that
$$\frac{\sqrt{\frac{ab + bc + ca}{3}}}{\sqrt{2a^2 + 5ab + 2b^2}}\ge \frac{8c^2 + 9(a + b)c + 8ab}{8(a^2 + b^2 + c^2) + 26(ab + bc + ca)}. \tag{1}$$
Note: Taking cyclic sum on (1), we get the desired inequality.
If $c = 0$, it is easy.
If $c > 0$, WLOG, assume that $c = 1$. Let $p = a + b, ~ q = ab$. Then $0 \le q \le p^2/4$. It suffices to prove that
$$\frac{\sqrt{(q + p)/3}}{\sqrt{2(p^2 - 2q) + 5q}}
\ge \frac{8 + 9p + 8q}{8(p^2 - 2q + 1) + 26(q + p)}.$$
Squaring both sides, it suffices to prove that
\begin{align*}
 &-92\,{q}^{3}+ \left( -224\,{p}^{2}+188\,p-224 \right) {q}^{2}+ \left( 
 64\,{p}^{4}-288\,{p}^{3}+313\,{p}^{2}+144\,p-128 \right) q \\
 &\quad +64\,{p}^{5}
 -70\,{p}^{4}-60\,{p}^{3}+32\,{p}^{2}+64\,p \ge 0.
\end{align*}
Denote LHS by $f(q)$.
We have $f''(q) = - 448p^2 + 376p - 448 - 552q$.
So, we have $f''(q) < 0$ on $q \ge 0$.
So, $f(q)$ is concave on $q \ge 0$.
Also, we have $f(0) = 64p^5 - 70p^4 - 60p^3 + 32p^2 + 64p \ge 0$
and $f(p^2/4) = \frac{1}{16}\,p \left( 9\,{p}^{3}+96\,{p}^{2}+256\,p+256 \right)  \left( p-2
\right) ^{2} \ge 0$.
Thus, $f(q) \ge 0$ for all $q\in [0, p^2/4]$.
We are done.
A: 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 !
