# Find the maximum value of the product $\ x_1^2 x_2^2\cdots x_n^2$

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

• – StubbornAtom May 19 '18 at 19:11

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 :

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

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