Inequality : $ \sqrt{\frac{a}{b+1}} + \sqrt{\frac{b}{c+1}} +\sqrt{\frac{c}{a+1}} \le \frac{3}{2} $ $a+b+c =1$ ,  $a, b, c>0 $ 
Prove
$$ \sqrt{\frac{a}{b+1}} + \sqrt{\frac{b}{c+1}} + \sqrt{\frac{c}{a+1}} \le  \frac{3}{2}$$
When I meet the inequality with square root symbols, I don't have any idea. It's my tragedy. I know the triangular inequality : $$\cos A + \cos B + \cos C  \le \frac{3}{2}  $$ 
But I failed to find a link between the two inequalities. I want some another hints. Thank you.

In eliminating square roots, a well chosen CS is your friend, (and for higher roots, often Holder).  Note by Cauchy-Schwarz inequality,
$$\sum_{cyc} (a+1) \cdot\sum_{cyc} \frac{a}{(a+1)(b+1)}\geqslant \left(\sqrt{\frac{a}{b+1}}+\sqrt{\frac{b}{c+1}}+\sqrt{\frac{c}{a+1}} \right)^2$$
Hence it is enough to show that
$$\sum_{cyc} \frac{a}{(a+1)(b+1)} \leqslant \frac9{16}$$
$$\iff 16\sum_{cyc} a(c+1) \leqslant 9\prod_{cyc} (a+1)$$
$$\iff 9abc+2 \geqslant 7(ab+bc+ca)$$
In symmetric form this is:
$$2(a+b+c)^3 + 9abc \geqslant 7(a+b+c)(ab+bc+ca)$$
$$\iff 2(a^3+b^3+c^3) \geqslant \sum_{cyc} ab(a+b)$$
which is true by Muirhead’s theorem.
