Do a;b;c>0 nên:
[tex]\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}> \frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\left ( 1 \right )[/tex]
[tex]\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< \frac{a+c}{a+b+c}+\frac{b+a}{a+b+c}+\frac{c+b}{a+b+c}=2\left ( 2 \right )[/tex]
Từ (1) và (2) =>đpcm