Giải thích các bước giải:
Ta có:
$P=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{abc}$
$\to P=\dfrac{1}{a^2+b^2+c^2}+\dfrac{a+b+c}{abc}, (a+b+c=1)$
$\to P=\dfrac{1}{a^2+b^2+c^2}+\dfrac1{ab}+\dfrac{1}{bc}+\dfrac1{ca}$
$\to P\ge \dfrac{1}{a^2+b^2+c^2}+\dfrac9{ab+bc+ca}$
$\to P\ge \dfrac{1}{a^2+b^2+c^2}+\dfrac1{ab+bc+ca}+\dfrac1{ab+bc+ca}+\dfrac7{ab+bc+ca}$
$\to P\ge \dfrac{9}{a^2+b^2+c^2+ab+bc+ca+ab+bc+ca}+\dfrac7{\dfrac{(a+b+c)^2}{3}}$
$\to P\ge \dfrac{9}{(a+b+c)^2}+\dfrac7{\dfrac{(a+b+c)^2}{3}}$
$\to P\ge 30$
Dấu = xảy ra khi $a=b=c=\dfrac13$
