If $a$, $b$ and $c$ are positive then $\sum\limits_{cyc} \frac{a^{2}}{b^{2}+c^{2}+bc}\geq 1$.
I tried to solve this problem by C-S. But I can't sovle it.
Things I have done so far:
$\sum\limits_{cyc} \frac{a^{2}}{b^{2}+c^{2}+bc}\geq \frac{(\sum\limits_{cyc}a)^2}{2.\sum\limits_{cyc}a^2+\sum\limits_{cyc}ab}\geq \frac{(\sum\limits_{cyc}a)^2}{3.\sum\limits_{cyc}a^2}=\frac{\sum\limits_{cyc}a^2+2.\sum\limits_{cyc}ab}{3.\sum\limits_{cyc}a^2}=\frac{1}{3}+\frac{2\sum\limits_{cyc}ab}{3.\sum\limits_{cyc}a^2}\geq ?$
Best Answer
By C-S and Muirhead $$\sum_{cyc}\frac{a^2}{b^2+c^2+bc}=\sum_{cyc}\frac{a^4}{a^2b^2+a^2c^2+a^2bc}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^2b^2+a^2c^2+a^2bc)}=$$ $$=\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^2b^2+a^2bc)}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^2b^2+a^4)}=1.$$ Also, by Holder and AM-GM: $$\sum_{cyc}\frac{a^2}{b^2+c^2+bc}=\sum_{cyc}\frac{a^3}{b^2a+c^2a+abc}\geq\frac{(a+b+c)^3}{3\sum\limits_{cyc}(ab^2+ac^2+abc)}=$$ $$=\frac{(a+b+c)^3}{\sum\limits_{cyc}(abc+3a^2b+3a^2c+2abc)}\geq\frac{(a+b+c)^3}{\sum\limits_{cyc}(a^3+3a^2b+3a^2c+2abc)}=1.$$