ta có $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{3}{abc}$cần chứng minh $\frac{3}{abc}\geq a^2+b^2+c^2 \Leftrightarrow abc(a+b+c)\leq 3$
bổ đề $(ab+bc+ca)^2\geq 3abc(a+b+c) \Rightarrow abc\leq \frac{(ab+bc+ca)^2}{9}$
$\Rightarrow abc(a^2+b^2+c^2)\leq \frac{(ab+bc+ca)^2(a^2+b^2+c^2)}{9}\leq \frac{1}{9}\left ( \frac{(a+b+c)^2}{3} \right )^3=3(cauchy)$
=> đpcm