Ta có $\frac{\sqrt a+\sqrt b+\sqrt c}{\sqrt 2} \overset{C-S}\le \frac{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}}{2}$
Nên chỉ cần cm $\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}} \ge \frac{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}}{2}$
$\leadsto \sum\left(\frac{a}{\sqrt{b+c}}-\frac{\sqrt{b+c}}{2}\right) \ge0$
$\leadsto \sum\frac{(a-b)+(a-c)}{2\sqrt{b+c}} \ge0$
$\leadsto (a-b)\left( \tfrac{1}{\sqrt{b+c}}-\tfrac{1}{\sqrt{c+a}}\right)+(b-c)\left(\tfrac{1}{\sqrt{c+a}}-\tfrac{1}{\sqrt{a+b}} \right)+(c-a)\left(\tfrac{1}{\sqrt{a+b}}-\tfrac{1}{\sqrt{b+c}} \right) \ge0$
KMTTQ,giả sử $a\ge b \ge c$, dễ dàng cm$\; (c-a)\left(\tfrac{1}{\sqrt{a+b}}-\tfrac{1}{\sqrt{b+c}} \right) \ge0$
Khi đó chỉ cần cm
$(a-b)\left( \tfrac{1}{\sqrt{b+c}}-\tfrac{1}{\sqrt{c+a}}\right)+(b-c)\left(\tfrac{1}{\sqrt{c+a}}-\tfrac{1}{\sqrt{a+b}} \right) \ge0$
$\leftrightsquigarrow \frac{(a-b)(\sqrt{c+a}-\sqrt{b+c})}{\sqrt{(b+c)(c+a)}}+\frac{(b-c)(\sqrt{a+b}-\sqrt{a+c})}{\sqrt{(c+a)(a+b)}} \ge0$
$\leftrightsquigarrow\frac{(a-b)^2}{(\sqrt{c+a}+\sqrt{b+c})\sqrt{b+c}}+\frac{(b-c)(a-c)}{(\sqrt{a+b}+\sqrt{b+c})\sqrt{a+b}} \ge0$
BDT cuối luôn đúng suy ra dpcm