Ta có:
$\dfrac{a+c}{a+b}+\dfrac{b+d}{b+c}+\dfrac{c+a}{c+d}+\dfrac{d+b}{d+a}$
$=(a+c)\left(\dfrac{1}{a+b}+\dfrac{1}{c+d}\right)+(b+d)\left(\dfrac{1}{b+c}+\dfrac{1}{d+a}\right)$
$\ge(a+c).\dfrac{4}{a+b+c+d}+(b+d).\dfrac{4}{a+b+c+d}=4$.
Dấu bắng xảy ra khi $a=c;b=d$