Ta có: \(F=\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\)
\(\Leftrightarrow F=\frac{a^2}{ab+ac}+\frac{b^2}{bc+bd}+\frac{c^2}{cd+ca}+\frac{d^2}{da+db}\)
\(\Leftrightarrow F\ge\frac{\left(a+b+c+d\right)^2}{2ac+2bd+\left(a+c\right)\left(b+d\right)}=P\)
\(\Leftrightarrow P=\frac{a^2+b^2+c^2+2ab+2bc+2cd+2ad+2ac+2bd}{ab+ac+bc+bd+cd+ac+ad+bd}\)
\(\Leftrightarrow P=\frac{\left(a^2+c^2\right)+\left(b^2+d^2\right)+2ab+2bc+2cd+2ad+2ac+2bd}{2ac+2bd+ab+bc+cd+ad}\)
(Vì \(a^2+c^2\ge2ac\Leftrightarrow\left(a-c\right)^2\ge0\)luôn đúng; \(b^2+d^2\ge2bd\Leftrightarrow\left(b-d\right)^2\ge0\)luôn đúng)
\(\Leftrightarrow P\ge\frac{2ac+2bd+2ab+2bc+2cd+2ad+2ac+2bd}{2ac+2bd+ab+cd+ad+ac+bd}\)
\(\Leftrightarrow P\ge\frac{4ac+4bd+2ab+2bc+2cd+2ad}{2ac+2bd+ab+bc+cd+ad}=2\)
\(\Leftrightarrow F\ge P\ge2\)
\(\LeftrightarrowĐPCM\)
