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.
 A: This is too long for a comment but it's more of a suggestion than an answer. Without loss of generality let $a \leq b \leq c. $ Put  $a = x,\ b = y,\ c = z = 1 - x - y$:
Let $$w(x,y) = \frac{x}{\sqrt{x+2y}} + \frac{y}{\sqrt{y+2(1-x-y)}} + \frac{1-x-y}{\sqrt{1-x-y+2x}}  $$
The constraint $a + b + c = 1$ means that x can be no larger then 1/3, otherwise y must be greater than 1/3, and z is forced to be less than 1/3 contrary to assumption. So $0 < x \leq 1/3.$ If x is minimally (almost) $0$, y is at most $1/2$ otherwise z is less than y contrary to assumption. 
So $x\leq y \leq 1/2.$
So the problem is now: 
Maximize w(x,y)
subject to
$0 < x\leq 1/3$
$x\leq y \leq 1/2$                                 
If the maximum we find is less than $\sqrt{\frac{3}{2}}$ the original inequality is true.  
We should verify (formally) the visual evidence of a 3D plot of the feasible region, which is that in $\{0 < x\leq 1/3, x\leq y \leq 1/2\}$ the function $w(x,y)$ has a local maximum on $y = 0^+$ (a last reminder that $0<x,y$) for suitable choice of x. Then to find that maximum we can let y equal $0$ and set the derivative $w_x(x,0) $ equal to zero. This (using a numerical routine) gives $x\approx 0.1547.$ A plot of $w(x, 0)$ shows this to be a maximum at about 1.179. 
The value of $\sqrt{3/2}$ is about 1.22, so $w(x,y)\ll \sqrt{3/2}$ for this point.
Again, using the visual shortcut, there also appears to be a local maximum for $w(x,y)$ on $x=0$ for suitable choice of y. If we let $w_y(0,y)= 0$ we (again numerically) obtain a value of $y \approx 0.845,$ at which $w(0,y)$ appears to be a maximum, but this is outside the feasible region, and the maximum is attained in the feasible region at $w(0,1/2) = 1.115,$ which is less than 1.179 and not maximal for the region as a whole.
So this sketch of an argument, which omits some important formalities, suggests the inequality is true. 
Edit: Khue noted a problem with this answer and suggested a simple fix. We cannot without generality assume $x < y < z,$ but can assume $x =\text{ min}(x,y,z),$ then the problem is minimize $w(x,y)$ subject to: 
$0\leq x \leq 1/2$, and $x \leq y \leq 1.$
As Khue notes, this changes the feasible region. Again we can plot the feasible region and the max appears to occur along $x = 0, 0 \leq y \leq 1.$ As before finding $w_y(0,y)$ the max occurs at about $y = 0.845299$ and at that value $w = 1.17996.$
A: 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!
A: We have over $(a,b,c)$:
$$\displaystyle LHS := \sum_{cyc} \frac{a}{\sqrt{a+2b}} = \frac{\sum_{cyc}\sqrt{a^2(b+2c)(c+2a)}}{\sqrt{(a+2b)(b+2c)(c+2a)}}$$
Using CS:
$$\displaystyle LHS \leq \sqrt{\frac{\left(a^2(b+2c)+b^2(c+2a)+c^2(a+2b)\right)\left(3(a+b+c)\right)}{(a+2b)(b+2c)(c+2a)}} \\ 
= \sqrt{3}\sqrt{\frac{a^2(b+2c)+b^2(c+2a)+c^2(a+2b)}{(a+2b)(b+2c)(c+2a)}} \\
= \sqrt{\frac{3}{2}}\sqrt{1-\frac{9abc}{2(a+2b)(b+2c)(c+2a)}} < \sqrt{\frac{3}{2}}$$
