Ta có : $\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\geq \frac{9}{ab+ac+bc}$
$\frac{1}{a^2+b^2+c^2}+\frac{1}{(ab+ac+bc)}+\frac{1}{(ab+ac+bc)} \geq\frac{9}{a^2+b^2+c^2+2ab+2ac+2bc}= \frac{9}{(a+b+c)^2}=9$ (*)
$\frac{21}{3(ab+ac+bc)}\geq \frac{21}{(a+b+c)^2}=21$ (**)
Cộng (*)+(**) => $P\geq 21+9=30$
dấu = xảy ra khi $a=b=c=\frac{1}{3}$
- congtuholi yêu thích