This question was asked in a previous year paper for a masters exam for which I am preparing and unfortunately I have no clue on which result should I use to solve it.
Let E={ $(x,y,z) \in \mathbb{R}^3 | x,y,z >0 , xy+yz+zx =1$}. Prove that there exists (a,b,c) $\in E$ such that $abc \geq xyz$ for all (x,y,z)$\in E$.
so, I have to prove that there exists a maxima a,b,c belonging to E.
I tried using the method of lagrange multipliers but I was unable to prove it. I think there must be some other methods.
So,can you please help.
I have studied real analysis from Tom M Apostol and Complex analysis from Ponnusamy and silvermann
It's an application of the arithmetic mean/geometric mean inequality.
The arithmetic mean of $xy$, $yz$, and $zx$ is $\frac{1}{3}$.
The geometric mean of the three will be $\sqrt[3]{x^2 y^2 z^2} = (x y z)^{2/3}$.
So we know that $(xyz)^{2/3} \leq \frac{1}{3}$, hence $xyz \leq \frac{1}{3^{3/2}}$.
And equality is achieved when $x = y = z = \frac{1}{\sqrt{3}}$.