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 :


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 :

$$x_N^2\prod ^{N-1} _i x_i^2 = \frac {x_N^2 (1-x_N^2)^{N-1}} {(N-1)^{N-1}}$$

  • $\begingroup$ Ok, but how can I prove that the maximum product is when they are the same? @StephenG $\endgroup$ – CocoaLapin Jun 12 '17 at 11:09

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