Giải thích các bước giải:
Ta có :
$ab+bc+ca\ge 3\sqrt[3]{ab.bc.ca}\to \dfrac{(ab+bc+ca)^3}{3^3}\ge a^2b^2c^2$
Mà $ab+bc+ca\le \dfrac{(a+b+c)^2}{3}=\dfrac 13$
$\to a^2b^2c^2\le \dfrac{(ab+bc+ca)^2}{3^4}$
$\to abc\le \dfrac{ab+bc+ca}{3^2}$
$\to P\ge \dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ca}$
$\to P\ge \dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{2ab+2bc+2ca}+\dfrac{17}{2(ab+bc+ca)}$
$\to P\ge \dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{7}{ab+bc+ca}$
$\to P\ge \dfrac{9}{a^2+b^2+c^2+ab+bc+ca+ab+bc+ca}+\dfrac{7}{ab+bc+ca}$
$\to P\ge \dfrac{9}{(a+b+c)^2}+\dfrac{7}{ab+bc+ca}$
$\to P\ge \dfrac{9}{1^2}+\dfrac{7}{\dfrac{1}{3}}$
$\to P\ge 30$
Dấu = xảy ra khi $a=b=c=\dfrac 13$
