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### Inequality

Posted: Sun Aug 13, 2017 4:55 pm
Let $x, y,z >0$ satisfying $x+y+z=1$. Prove that

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \sqrt{\frac{3}{xyz}}$

### Re: Inequality

Posted: Mon Aug 14, 2017 9:47 am
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, $$\displaystyle{x=y=z=\dfrac{1}{3}}$$.

### Re: Inequality

Posted: Sun Aug 20, 2017 9:25 am
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 [WeitzenbĂ¶ck]

\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

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \sqrt{\frac{3}{xyz}}$