inequality for positive number $a+b+c=3$

Question

Let $a,b,c$ be positive numbers such that $a+b+c=3$. Prove that

$$\frac{2-\sqrt{a}}{\sqrt{c+3a}}+\frac{2-\sqrt{b}}{\sqrt{a+3b}}+\frac{2-\sqrt{c}}{\sqrt{b+3c}}\:\ge \:\frac{3}{2}.$$

Answer 1

By AM-GM and C-S we obtain: $$\sum_{cyc}\frac{2-\sqrt{a}}{\sqrt{c+3a}}=\sum_{cyc}\frac{2(4-2\sqrt{a})}{2\sqrt{4(c+3a)}}\geq\sum_{cyc}\frac{2(4-a-1)}{4+c+3a}=$$ $$=\sum_{cyc}\frac{6(b+c)}{4(a+b+c)+3c+9a}=\sum_{cyc}\frac{6(b+c)}{13a+4b+7c}=$$ $$=\sum_{cyc}\frac{6(b+c)^2}{(b+c)(13a+4b+7c)}\geq\frac{6\left(\sum\limits_{cyc}(b+c)\right)^2}{\sum\limits_{cyc}(b+c)(13a+4b+7c)}=$$ $$=\frac{24(a+b+c)^2}{\sum\limits_{cyc}(11a^2+37ab)}\geq\frac{24(a+b+c)^2}{\sum\limits_{cyc}(16a^2+32ab)}=\frac{3}{2}.$$ Done!