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Let $a,b,c$ be positive reals, such that $\frac1a+\frac1b+\frac1c=a+b+c\ (\star)$, find the maximum of $\frac{1}{(2a+b+c)^2}+\frac{1}{(2b+c+a)^2}+\frac{1}{(2c+a+b)^2}$

This should be an application of Jensen's Inequality, so I have to find a concave function to maximize the sum, I thought to define;

$f(\frac1x):=\frac{1}{x+a+b+c},\quad\frac{\partial^2f(x)}{\partial x^2}<0$

so $f$ is concave and then the sum above is;

$f(\frac1a)+f(\frac1b)+f(\frac1c)$

using Jensen it should be less or equal to

$3\cdot f(\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}3)=3\frac{1}{\left(\frac3{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}+a+b+c\right)^2}\overset{(\star)}=3\frac{1}{\left(\frac3{a+b+c}+a+b+c\right)^2}$

and the denominator can be minimized with AM-GM;

$\left(\frac3{a+b+c}+a+b+c\right)^2\ge\left(2\sqrt{3\frac{a+b+c}{a+b+c}}\right)^2=12$

so the result is $\frac14$, but it is wrong, where does it fail, maybe AM-GM value cannot be attained ?

ketum
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1 Answers1

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I would rather show the maximum is $\frac3{16}$ by getting rid of the cumbersome constraint by homogenizing to: $$\frac{abc(a+b+c)}{ab+bc+ca}\sum_{cyc} \frac1{(2a+b+c)^2} \le \frac3{16}$$ So if $a+b+c=3$, this is equivalent to $$16\sum_{cyc} \frac1{(3+a)^2} \le \frac1a+\frac1b+\frac1c$$ which follows from $f(a)+f(b)+f(c)\ge 0$ with $$f(x) = \frac1x-\frac{16}{(3+x)^2}+\frac12(x-1) = \frac{(x-1)^2(x^2+7x+18)}{2x(3+x)^2} \ge 0$$

Macavity
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  • It seems that we can show that $a+b+c\ge 3$, comparing Equality $(\star)$ with AM-HM:

    $$\frac{a+b+c}3\ge \frac 3{\frac 1a+\frac 1a+\frac 1a}=\frac 3{a+b+c}$$ or applying Jensen’s inequality to a convex function $g(t)=1/t-t$:

    $$0=\frac{g(a)+g(b)+g(c)}3\ge g\left(\frac{a+b+c}3\right)=\frac 3{a+b+c}-\frac{a+b+c}3.$$

     – Alex Ravsky Jun 04 '15 at 12:01
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    @AlexRavsky Before homogenizing, it is true that $a+b+c = \frac1a+\frac1b+\frac1c \ge 3$. After homogenizing, that can be thrown out of the window, and we are free to choose other constraints as desired. – Macavity Jun 04 '15 at 12:03
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    @AlexRavsky Yes also by Titus lemma it can be shown $(\frac{1^2}a+\frac{1^2}b+\frac{1^2}c)(a+b+c)\ge(1+1+1)^2$ – ketum Jun 04 '15 at 12:03