Question

 Cho a, b, c, d > 0. Chứng minh:

\(\frac{a}{b+a}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge2\)

Answer 1

Áp dụng BĐT Cauchy Schwarz dạng Engel và BĐT AM - GM ta có :

\(M=\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\)

\(=\frac{a^2}{ab+ac}+\frac{b^2}{bc+bd}+\frac{c^2}{cd+ac}+\frac{d^2}{ad+bd}\)

\(\ge\frac{\left(a+b+c+d\right)^2}{ad+bc+cd+ab+2ac+2bd}\)

\(=\frac{2\left(a+b+c+d\right)^2}{\left(2ad+2bc+2cd+2ab+2ac+2bd\right)+2ac+2bd}\)

\(\ge\frac{2\left(a+b+c+d\right)^2}{\left(2ad+2bc+2cd+2ab+2ac+2bd\right)+a^2+b^2+c^2+^2}\)

\(=\frac{2\left(a+b+c+d\right)^2}{\left(a+b+c+d\right)^2}=2\)

Dấu "=" xảy ra khi a = b = c = d

Chúc bạn học tốt !!!