Symmetric and homogeneous three variable inequality with radicals. While trying to solve a problem, I got the following inequality which appears correct, but I cannot prove.  For positive $x, y, z$,
$$\sum_{cyc} \frac{x}{y^2+z^2} \ge \sum_{cyc} \sqrt{\frac{y^2+z^2}{2(x^2+y^2)(z^2+x^2)}}$$
Equality is obviously when $x=y=z$.  Tried AM-GM (takes it in opposite directions), Cauchy Schwarz (not tight enough) etc. with no luck. Adding constraints doesn't seem to simplify either.  Any hints welcome.

By C-S $\left(\sum\limits_{cyc}\sqrt{\frac{y^2+z^2}{2(x^2+y^2)(x^2+z^2)}}\right)^2\leq\sum_(y^2+z^2)\sum\limits_{cyc}\frac{1}{2(x^2+y^2)(x^2+z^2)}=\frac{2(x^2+y^2+z^2)^2}{\prod\limits_{cyc}(x^2+y^2)}$.
Thus, it remains to prove that
$\left(\sum\limits_{cyc}(x^5+x^3y^2+x^3z^2+x^2y^2z)\right)^2\geq2(x^2+y^2+z^2)^2(x^2+y^2)(x^2+z^2)(y^2+z^2)$.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$. 
Hence, we need to prove that $f(w^3)\geq0$, where
$f(w^3)=(81u^5-135u^3v^2+54uv^4+9u^2w^3-5v^2w^3)^2-$
$-2(3u^2-2v^2)^2(81u^2v^4-54v^6-54u^3w^3+36uv^2w^3-w^6)$.
We see that $f'(w^3)=2(81u^5-135u^3v^2+54uv^4+9u^2w^3-5v^2w^3)(9u^2-5v^2)+$
$+2(3u^2-2v^2)^2(54u^3-36uv^2+2w^3)\geq0$.
Id est, $f$ is an increasing function, which says that $f$ gets a minimal value for extremal value of $w^3$, which happens in two following cases only.


*

*$y=1$ and $z\rightarrow0^+$, which gives $(x^2+1)^2(x^4+2x^3+x^2+2x+1)(x-1)^2$;

*$y=z=1$, which gives $(x^2+1)^2(x^3+2x^2+x+8)x(x-1)^2\geq0$. Done!
