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## Inequality

General Mathematics
Riemann
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### Inequality

Let $x, y,z >0$ satisfying $x+y+z=1$. Prove that

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \sqrt{\frac{3}{xyz}}$
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Papapetros Vaggelis
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### Re: Inequality

Hi Riemann.

It is sufficient to prove that

$\displaystyle{a+b+c\geq \sqrt{3\,a\,b\,c}}$, where $\displaystyle{a\,,b\,,c>1}$ and $\displaystyle{a\,b+b\,c+c\,a=a\,b\,c}$.

So,

\displaystyle{\begin{aligned}a+b+c\geq \sqrt{3\,a\,b\,c}&\iff (a+b+c)^2\geq 3\,a\,b\,c\\&\iff (a^2+b^2+c^2)+2\,(a\,b+b\,c+c\,a)-3\,(a\,b+b\,c+c\,a)\geq 0\\&\iff (a^2+b^2+c^2)-(a\,b+b\,c+c\,a)\geq 0\\&\iff 2\,a^2+2\,b^2+2\,c^2-2\,a\,b-2\,b\,c-2\,c\,a\geq 0\\&\iff (a-b)^2+(b-c)^2+(c-a)^2\geq 0 \end{aligned}}

and the last one is true. The equality holds, if, and only, if, $\displaystyle{a=b=c}$ and then

$\displaystyle{a^2+a^2+a^2=a^3\iff a^3=3\,a^2\iff a=3=b=c}$.

We conclude that, if $\displaystyle{x\,,y\,,z>0}$ such that $\displaystyle{x+y+z=1}$, then

$\displaystyle{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\geq \sqrt{\dfrac{3}{x\,y\,z}}}$

and the equality holds if, and only if, $\displaystyle{\dfrac{1}{x}=\dfrac{1}{y}=\dfrac{1}{z}=3}$

or equivalently, if, and only if, $\displaystyle{x=y=z=\dfrac{1}{3}}$.
Riemann
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Posts: 169
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

### Re: Inequality

Thank you Papapetros Vaggelis. My solution is as follows.

Since $\frac{1}{x} \; , \; \frac{1}{y} \; , \; \frac{1}{z} >0$ then the numbers

$\sqrt{\frac{1}{x} + \frac{1}{y}} \; , \; \sqrt{\frac{1}{x} +\frac{1}{z}} \; , \; \sqrt{\frac{1}{y} + \frac{1}{z}}$

could be sides of a triangle. The area of this triangle is

$\mathcal{A} = \frac{1}{2} \sqrt{\frac{1}{xy} + \frac{1}{xz} + \frac{1}{yz}} = \frac{1}{2} \sqrt{\frac{x+z+y}{xyz}} = \frac{1}{2\sqrt{xyz}}$

However , in any triangle is holds that [Weitzenböck]

\begin{equation*} a^2+b^2+c^2 \geq 4 \mathcal{A} \sqrt{3} \end{equation*}

where $\mathcal{A}$ is the area of the triangle. Thus

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \sqrt{\frac{3}{xyz}}$
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$