# 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.

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@Chris'sister I tried to use your hint but I couldn't find a solution. Using what you say I obtained: $\displaystyle \frac{a}{\sqrt{a+2b}}+\frac{b}{\sqrt{b+2c}}+\frac{c}{\sqrt{c+2a}}\leq \sqrt{\frac{a}{a+2b}+\frac{b}{b+2c}+\frac{c}{c+2a}}$ and now I have to prove that : $\displaystyle \sqrt{\frac{a}{a+2b}+\frac{b}{b+2c}+\frac{c}{c+2a}} \lt \sqrt{\frac{3}{2}}$ Now, what can I do? Thanks for hint and for trying to help me :) –  Iuli Aug 31 '12 at 22:59
Eventually, you may resort to a calculus solution. For instance, you may express LHS in terms of $a$ and $b$ , then consider b fixed and compute the maximum of the function. It should work. –  Chris's sis Sep 1 '12 at 15:43
The inequalities are often hard and require much experience, hard work and a lot of research in order to successfully deal with them. –  Chris's sis Sep 1 '12 at 15:48

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 $w = 1.17996.$

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The inequality is not symmetric, so we cannot assume that $a\le b\le c$. –  Khue Apr 1 at 12:37
Hi @daniel. Your solution uses computer's help so I don't know (or cannot check) whether (or how) the assumption would affect the arguments. However, I know that for $a\le b\le c$ the inequality is weak, because if $a\le b\le c$ then $f(a,b,c) \le f(b,a,c)$ (this can easily be verified). In other words, if the inequality is true for $b\le a \le c$ (or $a\ge b \ge c$), then it is also true for $a\le b\le c$, but the reverse direction is not true. –  Khue Apr 2 at 21:44
One more remark: you found the maximum at $y=0^+$ and $x\approx 0.1547$, but these values clearly violate the constraint $x\leq y \leq 1/2$. By the way, you might be interested in the closed-form values of the numerical values that you found. Here they are: $$x=\frac{2+2\sqrt{3}}{3+2\sqrt{3}} \approx 0.845, \quad y =\frac{1}{3+2\sqrt{3}} \approx 0.1547$$ and the maximum value is $$\frac{2\sqrt{3}-2}{\sqrt{2\sqrt{3}}}+\sqrt{-1+\frac{2}{\sqrt{3}}} \approx 1.179.$$ These values were totally found by hand (+ a pen and a scrap paper :D). –  Khue Apr 2 at 22:05
As I said the inequality is not symmetric. It is only cyclic, not symmetric (clearly $f(a,b,c)\neq f(b,a,c)$), thus if we assume $a\le b\le c$ then generality is lost. Your arguments above even prove that. Indeed, according to your arguments, if $a\le b\le c$ then the maximum value of $f(a,b,c)$ is only $1.115$, i.e. $f(a,b,c)$ will never reach its true maximum value $1.179$, which can be achieved only in the case $b\le a\le c$ (or $c\le b\le a$ or $a\le c\le b$). ...(continued below)... –  Khue Apr 3 at 10:18
@daniel: Yeah I think it should be fine now. (Btw I will definitely post my solution, using Mixing Variables method, but maybe on next Friday.) –  Khue Apr 4 at 11:27
We have over $(a,b,c)$:
$$\displaystyle LHS := \sum_{cyc} \frac{a}{\sqrt{a+2b}} = \frac{\sum_{cyc}\sqrt{a^2(b+2c)(c+2a)}}{\sqrt{(a+2b)(b+2c)(c+2a)}}$$
$$\displaystyle LHS \leq \sqrt{\frac{\left(a^2(b+2c)+b^2(c+2a)+c^2(a+2b)\right)\left(3(a+b+c)\right)}{(a+2b)(b+2c)(c+2a)}} \\ = \sqrt{3}\sqrt{\frac{a^2(b+2c)+b^2(c+2a)+c^2(a+2b)}{(a+2b)(b+2c)(c+2a)}} \\ = \sqrt{\frac{3}{2}}\sqrt{1-\frac{9abc}{2(a+2b)(b+2c)(c+2a)}} < \sqrt{\frac{3}{2}}$$
Unfortunately, the first identity does not hold! For $a = b = c = 1/3$ we have $\displaystyle \sum_{cyc} \frac{a}{\sqrt{a+2b}} = 1 > \frac 1 {\sqrt 3} = \sqrt {\sum_{cyc} \frac{a^2}{a+2b}}$ –  AlbertH Sep 7 '12 at 21:53