Proving inequality $\frac{a}{\sqrt{a+2b}}+\frac{b}{\sqrt{b+2c}}+\frac{c}{\sqrt{c+2a}}\lt \sqrt{\frac{3}{2}}$ In the pdf which you can download here I found the following inequality which I can't solve it.
Exercise 2.1.10 Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that 
$$\displaystyle \frac{a}{\sqrt{a+2b}}+\frac{b}{\sqrt{b+2c}}+\frac{c}{\sqrt{c+2a}}\lt \sqrt{\frac{3}{2}} . $$
Thanks.

By C-S $$\left(\sum_{cyc}\frac{a}{\sqrt{a+2b}}\right)^2\leq\sum_{cyc}\frac{a}{(a+2b)(a+2c)}\sum_{cyc}a(a+2c)=\sum_{cyc}\frac{a}{(a+2b)(a+2c)}.$$
Thus, it remains to prove that
$$\sum_{cyc}\frac{a}{(a+2b)(a+2c)}<\frac{3}{2(a+b+c)}$$ or
$$\sum_{cyc}(4a^3b^3+4a^4bc+24a^3b^2c+24a^3c^2b+25a^2b^2c^2)>0.$$
Done!
