Giải thích các bước giải:

Ta có : $\left(x+y\right)^2\ge 4xy\to\dfrac{x+y}{xy}\ge \dfrac{4}{x+y}$

$\to \dfrac{4}{x+y}\le\dfrac{1}{x}+\dfrac1y$

$\to \dfrac{1}{x+y}\le\dfrac14\left(\dfrac{1}{x}+\dfrac1y\right)$

Áp dụng ta có :
$\dfrac{a}{b+c}=a.\dfrac1{b+c}\le a.\dfrac14\left(\dfrac1b+\dfrac1c\right)=\dfrac14\left(\dfrac ab+\dfrac ac\right)$

Tương tự

$\dfrac{b}{c+a}\le\dfrac14\left(\dfrac bc+\dfrac ba\right)$

$\dfrac{c}{a+b}\le\dfrac14\left(\dfrac ca+\dfrac cb\right)$

$\to \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\le\dfrac14\left(\dfrac ab+\dfrac ac\right)+ \dfrac14\left(\dfrac bc+\dfrac ba\right)+\dfrac14\left(\dfrac ca+\dfrac cb\right)$

$\to \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\le\dfrac14\left(\dfrac ab+\dfrac ac+\dfrac bc+\dfrac ba+\dfrac ca+\dfrac cb\right)$

$\to 4\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\le\dfrac ab+\dfrac ac+\dfrac bc+\dfrac ba+\dfrac ca+\dfrac cb$

$\to 4\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\le\left(\dfrac ab+\dfrac cb\right)+\left(\dfrac ac+\dfrac bc\right)+\left(\dfrac ba+\dfrac ca\right)$

$\to 4\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\le\dfrac{a+c}b+\dfrac{a+b}c+\dfrac{b+c}a$

$\to 4\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\le\dfrac{a+b}c+\dfrac{b+c}a+\dfrac{a+c}b$

Dấu = xảy ra khi $a=b=c$