Giải thích các bước giải:
Ta có :
$A=(\dfrac{a}{(ab+a+1)^2}+\dfrac{b}{(bc+b+1)^2}+\dfrac{c}{(ca+c+1)^2})(a+b+c)$
$\to A\ge (\sqrt{\dfrac{a}{(ab+a+1)^2}.a}+\sqrt{\dfrac{b}{(bc+b+1)^2}.b}+\sqrt{\dfrac{c}{(ca+c+1)^2}.c})^2$
$\to A\ge \dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ca+c+1}$
$\to A\ge \dfrac{a}{ab+a+1}+\dfrac{ab}{abc+ab+a}+\dfrac{abc}{abca+abc+ab}$
$\to A\ge \dfrac{a}{ab+a+1}+\dfrac{ab}{1+ab+a}+\dfrac{1}{a+1+ab}$
$\to A\ge 1$
$\to (\dfrac{a}{(ab+a+1)^2}+\dfrac{b}{(bc+b+1)^2}+\dfrac{c}{(ca+c+1)^2})(a+b+c)\ge 1$
$\to \dfrac{a}{(ab+a+1)^2}+\dfrac{b}{(bc+b+1)^2}+\dfrac{c}{(ca+c+1)^2}\ge \dfrac1{a+b+c}$
Dấu = xảy ra khi $a=b=c$
