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Inequality
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 \(\displaystyle{a\,b+b\,c+c\,a=a\,b\,c}\).
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
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
\[\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \sqrt{\frac{3}{xyz}}\]