Ineq!⠀⠀⠀

by cjquines0, Dec 29, 2016, 9:49 AM

MOP 2002 wrote:
Let $a, b, c$ be positive real numbers. Prove that $$\left(\frac{2a}{b+c}\right)^{2/3} + \left(\frac{2b}{c+a}\right)^{2/3} + \left(\frac{2c}{a+b}\right)^{2/3} \geq 3.$$

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Tidbit

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3 Comments

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Nice - Karamata's really is easy. :)

I'm definitely not an expert on what to include in a solution, but if I were to use Karamata's (see about four-ish posts down), I would definitely specify the convexity of my function just to clarify what direction my inequality is pointing.

Here's just to make sure your function is actually concave using derivative calculus.

by agbdmrbirdyface, Dec 29, 2016, 10:45 PM

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Hey, that's pretty good!

by zephyrcrush78, Dec 30, 2016, 1:32 AM

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Assume WLOG $a+b+c=1$ since the inequality is homogenous. We have to prove $$\sum_{cyc}a\cdot\sqrt[3]{\frac{1}{a(1-a)^2}}\ge\frac{3}{\sqrt[3]{4}}$$Consider function $$\forall_{x\in(0;1)}\ f(x)=\sqrt[3]{\frac{1}{x(1-x)^2}}$$Then $$\forall_{x\in(0;1)}\ f'(x)=\frac13\cdot(3x-1)\cdot x^{-4/3}\cdot (1-x)^{-5/3}$$and$$\forall_{x\in(0;1)}\ f''(x)=\frac{2\left((3x-1)^2+1\right)}{9x^{7/3}\cdot(1-x)^{8/3}}>0$$We have proven that $f(x)$ is a convex function. Thus
$$\sum_{cyc}a\cdot f(a)\ge f(s)$$where $$s=a^2+b^2+c^2$$Of course $0<ab+bc+ca\implies s<(a+b+c)^2=1$ so $f(s)$ exists in the given domain. We are left with proving $$\frac{1}{s(1-s)^2}\ge\frac{27}{4}$$That easily follows from AM-GM:
$$s\cdot \frac{1-s}{2}\cdot \frac{1-s}{2}\le\frac{1}{27}$$Awaiting for likes.

by WolfusA, Sep 9, 2019, 2:26 PM

To share with readers my favorite problem I came across today :) (Shout for contrib.)

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