Giải thích các bước giải:
Ta có :
$P=\dfrac{a^2(1-b)}{b}+\dfrac{b^2(1-c)}{c}+\dfrac{c^2(1-a)}{a}$
$\to P=\dfrac{a^2(1-b)^2}{b(1-b)}+\dfrac{b^2(1-c)^2}{c(1-c)}+\dfrac{c^2(1-a)^2}{a(1-a)}$
$\to P\ge\dfrac{(a(1-b)+b(1-c)+c(1-a))^2}{b(1-b)+c(1-c)+a(1-a)}$
$\to P\ge\dfrac{(a+b+c-(ab+bc+ca))^2}{a+b+c-(a^2+b^2+c^2)}$
$\to P\ge\dfrac{(a+b+c-(ab+bc+ca))^2}{a+b+c-(ab+bc+ca)}$
$\to P\ge a+b+c-(ab+bc+ca)$
$\to P\ge \sqrt{3(ab+bc+ca)}-(ab+bc+ca)$
$\to P\ge \sqrt{3}-1$
Dấu = xảy ra khi $a=b=c=\dfrac{1}{\sqrt{3}}$