<s>[INST] 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 $~\displaystyle\sum_\text{cyclic}\left(\dfrac{1}{y^{2}+z^{2}}+\dfrac{1}{1-yz}\right)\geq 9$
 [/INST] 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.
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