Find the maximum value of the product $$\ x_1^2 x_2^2\cdots x_n^2 $$ if $$\ x_1^2+x_2^2+\cdots+x_n^2=1 $$
I found out that, if $\ x_1^2=x_2^2=\cdots=x_n^2 $, then $\ x_1^2 x_2^2\cdots x_n^2 =1/n^n$ but I don't know how to prove that it's the biggest possible product
The arithmetic mean geometric mean inequality states that $$\frac{a_1+a_2+\cdots+a_n}n\ge\sqrt[n]{a_1a_2\cdots a_n} $$ whenever $a_1,\ldots,a_n\ge0$. Of course, equality holds when $a_1=\cdots=a_n$.
Can you find a useful substitution for the $a_j$ in terms of the $x_j$?
I don't know how to find the biggest product when $x_i$ are not the same.
But the maximum product is when they are the same, so you're actually asking a different question (kind of) when you fix one of the $x_i$ values.
What happens when you do ?
Let's arbitrarily pick $x_N$ to fix, in which case we want to maximize :
$$x_1^2x_2^2...x_{N-1}^2$$
We can ignore the $x_N^2$ as it's fixed and positive anyway.
Looks pretty similar to what we started with.
So what about the constraint ? That becomes :
$$\sum ^{N-1} _i x_i^2 = 1 - x_N^2$$
And as we can easily see that if we just define a new variable $y_i$
$$y_i := \frac {x_i} {\sqrt {1-x_N^2}}$$
We can now get our original problem form back in an $N-1$ dimensional form :
$$maximize\, \prod ^{N-1} _i y_i^2$$
$$subject\, to\, \sum ^ {N-1} _i y_i^2 = 1$$
Which is is the same as our original form and requires :
$$y_i = \frac 1 {\sqrt N}$$
and so :
$$x_N^2\prod ^{N-1} _i x_i^2 = \frac {x_N^2 (1-x_N^2)^{N-1}} {(N-1)^{N-1}}$$
