Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{bc+bd}+\frac{c^2}{cd+ca}+\frac{d^2}{da+db}\)
\(\geq \frac{(a+b+c+d)^2}{ab+ac+bc+bd+cd+ca+da+db}=\frac{(a+b+c+d)^2}{ab+cd+2ac+2bd+bc+da}\) (1)
Ta có:
\((a+b+c+d)^2=a^2+b^2+c^2+d^2+2ac+2bd+2(a+c)(b+d)\)
\(=a^2+b^2+c^2+d^2+2ac+2bd+2ab+2ad+2bc+2cd\)
Áp dụng BĐT AM-GM:
\(a^2+c^2\geq 2ac; b^2+d^2\geq 2bd\)
\(\Rightarrow (a+b+c+d)^2\geq 4ac+4bd+2ab+2ad+2bc+2cd\)
\(\Leftrightarrow (a+b+c+d)^2\geq 2(ab+cd+2ac+2bd+bc+da)\) (2)
Từ (1); (2) suy ra :
\(\text{VT}\geq \frac{2(ab+cd+2ac+2bd+bc+da)}{ab+cd+2ac+2bd+bc+da}=2\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=d\)
