Giải thích các bước giải:
Ta có :
$A=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}$
$\to A=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ba}+\dfrac{c^2}{ac+bc}$
$\to A\ge\dfrac{(a+b+c)^2}{ab+ac+bc+ba+ac+bc}$
$\to A\ge\dfrac{(a+b+c)^2}{2(ab+bc+ca)}$
$\to A\ge\dfrac{3(ab+bc+ca)}{2(ab+bc+ca)}$
$\to A\ge \dfrac 32$
$B=\dfrac{b+c}{a}+\dfrac{a+c}{b}+\dfrac{a+b}{c}$
$\to B=\dfrac ba+\dfrac ca+\dfrac ab+\dfrac cb+\dfrac ac+\dfrac ac$
$\to B\ge6\sqrt[6]{\dfrac ba.\dfrac ca.\dfrac ab.\dfrac cb.\dfrac ac.\dfrac ac}$
$\to B\ge 6$
$\to A+B\ge \dfrac 32+6=\dfrac{15}2$
$\to \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}+\dfrac{b+c}{a}+\dfrac{a+c}{b}+\dfrac{a+b}{c}\ge\dfrac{15}2$
Dấu = xảy ra khi $a=b=c$
