\(\frac{a}{a+b+c}>\frac{a}{a+b+c+d}\)
\(\frac{b}{b+c+d}>\frac{b}{a+b+c+d}\)
\(\frac{c}{c+d+a}>\frac{c}{a+b+c+d}\)
\(\frac{d}{d+a+b}>\frac{d}{a+b+c+d}\)
\(\Rightarrow\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}>\frac{a+b+c+d}{a+b+c+d}=1\left(1\right)\)
\(\frac{a}{a+b+c}< \frac{a+d}{a+b+c+d}\left(vì\frac{a}{a+b+c}< 1\right)\)
tương tự
\(\frac{b}{b+c+d}< \frac{b+a}{a+b+c+d}\)
\(\frac{c}{c+d+a}< \frac{c+b}{a+b+c+d}\)
\(\frac{d}{d+a+b}< \frac{d+c}{a+b+c+d}\)
\(\Rightarrow\)\(\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}< \frac{2.\left(a+b+c+d\right)}{a+b+c+d}=2\left(2\right)\)
từ (1) và (2) => đpcm
