Maximum of a $f(x)=(x_{1}\cdot x_{2}\cdot...\cdot x_{n})^{2}$ over the sphere $S_{n-1}$

Question

Yesterday I was looking in my Math Analysis book and I found this problem, which I'm very interested about cause I think it's not that easy. Can someone solve it? I tried but nothing came out of my mind.

Let $f$ be a function defined by: $$f(x)=(x_{1}\cdot x_{2}\cdot...\cdot x_{n})^{2}$$ and $S_{n-1}:=\{x\in\mathbb{R}^{n}$ : $\|x\|_{2}=1\}$.

(1) Show that $\underset{x\in S_{n-1}}{\max}f(x)=\frac{1}{n^{n}}.$

(2) Use the (1) to prove that $|(x_1\cdot x_2\cdot...\cdot x_n)|\leq\|x\|_{2}^{n}\cdot\frac{1}{n^{(n/2)}}$.

(3) Deduce from (1) that if $x=(x_{1},x_{2},...,x_{n})$ has only positive value-coordinates, so $$\frac{\sum_{i=1}^{n}x_{i}}{n}\geq\sqrt[n]{(x_{1}\cdot x_{2}\cdot...\cdot x_{n})}.$$

Thanks a lot :)

Answer 1

Hint. By AM-GM inequality, $$(f(x))^{1/n}=\left((x_{1}\cdot x_{2}\cdot\dots\cdot x_{n})^{2}\right)^{1/n}\leq \frac{x_{1}^2+ x_{2}^2+\dots+x_{n}^2}{n}=\frac{\|x\|_2^2}{n}.$$

Answer 2

One wants to say "By symmetry, all the $x_{i}$ must be equal". This simplification makes the problem trivial - the constraint becomes $nx_{i}^{2} = 1$, and the bound follows.

Since $S_{n-1}$ is a closed and bounded subset of $\mathbb{R}^{n}$, $f$ does attain a maximum somewhere. Let's try to prove that in order to maximise $f$, all the $x_{i}$ should be equal. Suppose $x_{i} \ne x_{j}$. Then the average $$\frac{x_{i}^{2}+x_{j}^{2}}{2}$$ is different from $x_{i}^{2}$ and $x_{j}^{2}$. Now suppose we replace $x_{i}$ and $x_{j}$ by $\bar{x} = \left((x_{i}^{2}+x_{j}^{2})/2\right)^{1/2}$ Then $||x||_{2}$ is still $1$, but we have the identity $$\left(\frac{x_{i}^{2}+x_{j}^{2}}{2}\right)^{2}=\left(\frac{x_{i}^{2}-x_{j}^{2}}{2}\right)^{2}+x_{i}^{2}x_{j}^{2}>x_{i}^{2}x_{j}^{2}$$ so the value of $f$ strictly increases! Therefore if $f$ is maximal, $x_{1}=x_{2} = \ldots =x_{n}$. Can you finish it from there?