# 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.

• The inequality toolbox consists of quite a lot of methods. But the first few to know would be "Sum of Squares", "AM-GM(-HM)" and "Cauchy-Schwarz". Do learn them and try to apply them here. (In my opinion, not objective fact) Commented Feb 3, 2018 at 14:28
• Thanks your words..I think that there are too many inequalities in the world. Commented Feb 3, 2018 at 14:52

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.
• @the4seasons Thanks. Instead of Muirhead, AM-GM suffices too, one just has to add several of $\frac23a^3+\frac13b^3\ge a^2b$. Commented Feb 4, 2018 at 4:03