Proving $~\sum_\text{cyclic}\left(\frac{1}{y^{2}+z^{2}}+\frac{1}{1-yz}\right)\geq 9$ $a$,$b$,$c$ are non-negative real numbers such that $~x^{2}+y^{2}+z^{2}=1$    
show that $~\displaystyle\sum_\text{cyclic}\left(\dfrac{1}{y^{2}+z^{2}}+\dfrac{1}{1-yz}\right)\geq 9$
 A: By C-S we obtain:
$$\sum_{cyc}\left(\frac{1}{y^2+z^2}+\frac{1}{1-yz}\right)=\sum_{cyc}\left(\frac{(3x+y+z)^2}{(3x+y+z)^2(1-x^2)}+\frac{(y+z)^2}{(y+z)^2(1-yz)}\right)\geq$$
$$\geq\frac{25(x+y+z)^2}{\sum\limits_{cyc}(3x+y+z)^2(1-x^2)}+\frac{4(x+y+z)^2}{\sum\limits_{cyc}(y+z)^2(1-yz)}.$$
Thus, it remains to prove that
$$\frac{25(x+y+z)^2}{\sum\limits_{cyc}(3x+y+z)^2(1-x^2)}+\frac{4(x+y+z)^2}{\sum\limits_{cyc}(y+z)^2(1-yz)}\geq9.$$
The last inequality we can prove by the $uvw$'s technique.
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, we need to prove that 
$$\frac{25u^2}{\sum\limits_{cyc}(3u+2x)^2(9u^2-6v^2-x^2)}+\frac{4u^2}{\sum\limits_{cyc}(3u-x)^2(9u^2-6v^2-yz)}\geq\frac{1}{9u^2-6v^2},$$
which  is eighth degree, which says that it's a quadratic inequality of $w^3$
and the rest is smooth, but a lot of work. 
The starting inequality is a known unsolved problem and you can be first, which ends a proof. 
Good luck! 
A: Letting $a = x^2, b = y^2, c = z^2$,
it suffices to prove that, for all $a, b, c > 0$ with $a + b + c = 1$,
$$\sum_{\mathrm{cyc}} \left(\frac{1}{b + c} + \frac{1}{1 - \sqrt{bc}}\right) \ge 9.$$
We have
$$\frac{1}{1 - \sqrt{bc}} = \frac{1 + \sqrt{bc}}{1 - bc}
\ge \frac{1 + \frac{2bc}{b + c}}{1 - bc} = \frac{b + c + 2bc}{(b+c)(1-bc)}.$$
It suffices to prove that
$$\sum_{\mathrm{cyc}} \left(\frac{1}{b + c} + \frac{b + c + 2bc}{(b+c)(1-bc)}\right) \ge 9$$
or
$$\sum_{\mathrm{cyc}} \frac{(1+b)(1+c)}{(b+c)(1-bc)} \ge 9.$$
The last inequality is killed by BW (Buffalo Way).
This proof is not nice. Hope to see a nice proof.
