Ta có:
$\sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{c+a}}$ $=2(\sqrt{\frac{1}{2}.\frac{a}{a+b}}+\sqrt{\frac{1}{2}.\frac{b}{b+c}}+\sqrt{\frac{1}{2}.\frac{c}{c+a}})$
$\le \frac{1}{2}+\frac{a}{a+b}+\frac{1}{2}+\frac{b}{b+c}+\frac{1}{2}+\frac{c}{c+a}$
$=\frac{9}{2}-(\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}) \le \frac{9}{2}-\frac{3}{2}=3\Rightarrow $ đpcm..
Đẳng thức xảy ra khi và chỉ khi $a=b=c.$