Ta có √a+√b+√c√2C−S≤√a+b+√b+c+√c+a2
Nên chỉ cần cm a√b+c+b√c+a+c√a+b≥√a+b+√b+c+√c+a2
\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