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Posted: Sun Aug 13, 2017 4:55 pm
by Riemann
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
by Papapetros Vaggelis
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 \(\displaystyle{a\,b+b\,c+c\,a=a\,b\,c}\).


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

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
by Riemann
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

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