# 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 wise sister 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 wise sister 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.

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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:

$$\displaystyle LHS \leq \sqrt{\frac{\left(a^2(b+2c)+b^2(c+2a)+c^2(a+2b)\right)\left(3(a+b+c)\right)}{(a+2b)(b+2c)(c+2a)}} \\ = \sqrt{3}\sqrt{\frac{a^2(b+2c)+b^2(c+2a)+c^2(a+2b)}{(a+2b)(b+2c)(c+2a)}} \\ = \sqrt{\frac{3}{2}}\sqrt{1-\frac{9abc}{2(a+2b)(b+2c)(c+2a)}} < \sqrt{\frac{3}{2}}$$

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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
Yes, you are right. The 1st eq I had incorrectly placed the \sqrt before the sum. The rest follows, however, and the edit should now make this clear. – cbyn Sep 10 '12 at 13:17