Question

[nhóc con]

với các số a,b,c,d dương hãy chứng minh

\(F=\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}>2\)

HOẶC BẰNG 2

Answer 1

[Nguyễn Linh Chi]

Áp dụng BĐT Svác - xơ.

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

\(=\frac{a^2}{ba+ca}+\frac{b^2}{cb+db}+\frac{c^2}{dc+ac}+\frac{d^2}{ad+bd}\)

\(\ge\frac{\left(a+b+c+d\right)^2}{ba+ca+bd+db+dc+ac+ad+bd}\)(1)

Xét:  \(\left(a+b+c+d\right)^2-2\left(ba+ca+bd+db+dc+ac+ad+bd\right)\)

\(=a^2+b^2+c^2+d^2-2bd-2ac\)

\(=\left(a-c\right)^2+\left(b-d\right)^2\ge0\)

=> \(\left(a+b+c+d\right)^2\ge2\left(ba+ca+bd+db+dc+ac+ad+bd\right)\)

=> \(\frac{\left(a+b+c+d\right)^2}{ba+ca+bd+db+dc+ac+ad+bd}\ge2\)(2)

Từ ( 1); (2) => \(F\ge2\)

Dấu "=" xảy ra <=> a = b = c = d.