$\dfrac{a}{b+c+d}+\dfrac{b}{a+c+d}+\dfrac{c}{d+a+b}+\dfrac{d}{a+b+c}=\dfrac{a^2}{ab+ac+ad}+\dfrac{b^2}{ab+bc+bd}+\dfrac{c^2}{cd+ac+bc}+\dfrac{d^2}{ad+bd+cd}\ge \frac{(a+b+c+d)^2}{2ab+2bc+2cd+2da+2ac+2bd}$
Ta sẽ chứng minh
$\frac{(a+b+c+d)^2}{2ab+2bc+2cd+2da+2ac+2bd} \ge\frac{4}{3}$
$\Leftrightarrow 3(a+b+c+d)^2 \ge 8\left ( ab+bc+cd+da+ac+bd \right )$
$\Leftrightarrow 3(a^2+b^2+c^2+d^2) -2\left ( ab+bc+cd+da+ac+bd \right ) \ge 0 $
$\Leftrightarrow (a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2+(a-c)^2+(b-d)^2 \ge 0 $, hiển nhiên đúng.