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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 :)2012-08-31
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    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.2012-09-01
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    The inequalities are often hard and require much experience, hard work and a lot of research in order to successfully deal with them.2012-09-01

3 Answers 3

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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) 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 $a\le b\le c$.2014-04-01
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    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.2014-04-02
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    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).2014-04-02
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    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)...2014-04-03
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    We can, however, assume that $a=\min(a,b,c)$ for example. In this case, the problem becomes maximizing $w(x,y)$ subject to $0 and $x\le y \le 1$. The feasible region has been changed a bit, but probably the arguments in your solution can still be applied. With this (necessary) modification, your solution becomes correct. (P/s: Of course a nice analytic solution exists, maybe I will post it when I have time.)2014-04-03
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    Hi @daniel. Do you realize that I have suggested a small modification to your solution to make it become correct? If you agree with my suggestion, then just modify it that way, and there's no need to delete the answer. If you don't, or if I was not clear enough, then just ask me for clarification. Best regards.2014-04-03
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    @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.)2014-04-04
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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}}$2012-09-07
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    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.2012-09-10
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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!