Bài 119(Thailand MO): Cho a,b,c>0. CMR:
$\frac{a^2}{b^2+bc+c^2}+\frac{b^2}{c^2+ca+a^2}+\frac{c^2}{a^2+ab+b^2}\geq 1$
Ta có: $2(b^2+bc+c^2)\leq 3(b^2+c^2)$
$\Leftrightarrow b^2+bc+c^2\leq\frac{3}{2}(b^2+c^2)$
$\Leftrightarrow\frac{a^2}{b^2+bc+c^2}\geq\frac{2a^2}{3(b^2+c^2)}$
$\Leftrightarrow\sum\frac{a^2}{b^2+bc+c^2}\geq\frac{2}{3}(\sum\frac{a^2}{b^2+c^2})\geq\frac{2}{3}.\frac{3}{2}=1$ (Bất đẳng thức $Nesbit$)