dựa vào tính chất hoán vị của a,b,c,d
Đặt \(a\ge b\ge c\ge d\)
tacó :
\(\frac{a}{b+c}\ge\frac{a}{a+c},\frac{c}{a+d}\ge\frac{c}{a+c}\)\(\frac{b}{c+d}\ge\frac{b}{b+d},\frac{d}{a+b}\ge\frac{d}{b+d}\)
=>\(\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{a+d}+\frac{d}{a+b}\ge\frac{a}{a+c}+\frac{b}{b+d}+\frac{c}{a+c}+\frac{d}{b+d}=\frac{a+c}{a+c}+\frac{b+d}{b+d}=2\)(ĐPCM)