Let $a, b, c$ be positive real numbers such that $a+b+c = 1$. Prove that $$ \displaystyle\sum_{cyc}\frac{ab}{\sqrt{ab+bc}} \leq \frac{1}{\sqrt{2}}$$
My attempted work :
By C-S, $$ (ab+ac)(1+1) \geq (\sqrt{ab}+\sqrt{bc})^2$$
$$\sqrt{2} \sqrt{ab+bc} \geq \sqrt{ab}+\sqrt{bc}$$
$$\frac{\sqrt{2} ab}{ \sqrt{ab}+\sqrt{bc}} \geq \frac{ab}{\sqrt{ab+bc}}$$
$$\frac{ab}{\sqrt{ab+bc}} \leq \frac{\sqrt{2} ab}{ \sqrt{ab}+\sqrt{bc}}$$
multiply through by $\sqrt{2}$
$$\displaystyle\sum_c \frac{\sqrt{2} ab}{\sqrt{ab+bc}} \leq \displaystyle\sum_c \frac{ 2ab}{ \sqrt{ab}+\sqrt{bc}} = \displaystyle\sum_c \frac{ ab}{ \sqrt{ab}+\sqrt{bc}} + \displaystyle\sum_c \frac{ bc}{ \sqrt{ab}+\sqrt{bc}} = \displaystyle\sum_c \frac{ ab+bc}{ \sqrt{ab}+\sqrt{bc}}$$
Please suggest, how to show that
$$\displaystyle\sum_c \frac{ ab+bc}{ \sqrt{ab}+\sqrt{bc}} \leq \frac{1}{\sqrt{2}}\sqrt{2} = 1 = a+b+c $$
Can we just use basic inequalities ?