Giải thích các bước giải:
Ta có:
$A=\dfrac{a}{a+4b+4c}+\dfrac{b}{b+4c+4a}+\dfrac{c}{c+4a+4b}$
$\to A=\dfrac{a^2}{a^2+4ab+4ca}+\dfrac{b^2}{b^2+4bc+4ab}+\dfrac{c^2}{c^2+4ca+4bc}$
$\to A\ge \dfrac{(a+b+c)^2}{a^2+4ab+4ca+b^2+4bc+4ab+c^2+4ca+4bc}$
$\to A\ge \dfrac{(a+b+c)^2}{a^2+b^2+c^2+2(ab+bc+ca)+6(ab+bc+ca)}$
$\to A\ge \dfrac{(a+b+c)^2}{(a+b+c)^2+2\cdot 3(ab+bc+ca)}$
$\to A\ge \dfrac{(a+b+c)^2}{(a+b+c)^2+2\cdot (a+b+c)^2}$
$\to A\ge \dfrac{(a+b+c)^2}{3\cdot (a+b+c)^2}$
$\to A\ge\dfrac13$
$\to đpcm$
