Giải thích các bước giải:
$P=\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$
$=\dfrac{a^2}{ab+ca-a^2}+\dfrac{b^2}{bc+ab-b^2}+\dfrac{c^2}{ac+bc-c^2}$
$\ge \dfrac{(a+b+c)^2}{ab+ca-a^2+bc+ab-b^2+ac+bc-c^2}$
$\ge \dfrac{(a+b+c)^2}{2(ab+bc+ca)-(a^2+b^2+c^2)}$
$\ge \dfrac{(a+b+c)^2}{\dfrac{2}{3}(a+b+c)^2-\dfrac{1}{3}(a+b+c)^2}=3$