Cyclic inequality with square roots and three variables

Question

Is the following inequality true for any a,b,c>0? I couldn't find a counterexample but I couldn't prove it either.

$\sum \sqrt{a+b} \ge \sqrt{2} \sum \sqrt{a}$

(The sum is cyclic)

Answer 1

$$\sqrt{a+b} \ge \frac{\sqrt{2}}{2} (\sqrt{a}+\sqrt{b}) $$ $$a+b \ge \frac{a+2\sqrt{ab}+b}{2}$$ $$a-2\sqrt{ab}+b \ge 0$$ $$(\sqrt{a}-\sqrt{b})^2 \ge 0$$ which is obviously true

Summing the inequality cyclically gets us to the desired inequality with equality for a=b=c

Answer 2

Also, by C-S: $$\sum_{cyc}\sqrt{a+b}=\frac{1}{\sqrt2}\sum_{cyc}\sqrt{(1^2+1^2)\left((\sqrt{a})^2+\sqrt{b})^2\right)}\geq\frac{1}{\sqrt2}\sum_{cyc}(\sqrt{a}+\sqrt{b})=\sum_{cyc}\sqrt{2a}.$$