Question

Giúp với mai thi rồi : ​

Cho a,b,c,d dương . CMR : \(\dfrac{a}{b+c}+\dfrac{b}{c+d}+\dfrac{c}{d+a}+\dfrac{d}{a+b}\ge2\)

Answer 1

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 \(a=b=c=d\)