With $\vec{x}=(x_1,\ldots,x_n)$, find all the min and max of $\prod_{i=1}^{n} x_{i}^i$ given that $||\vec{x}||=1$ With $\vec{x}=(x_1,\ldots,x_n)$,  find all the min and max of $\prod_{i=1}^{n} x_{i}^i$ given that $||\vec{x}||=1$
Now clearly this is Lagrange multiplier.
So one might take $\prod_{i=1}^{n} x_{i}^i-\lambda(\sum_{i=1}^{n}x^2_i -1)$. But the problem is when I take the derivative I get an algebraic mess.
Clearly the solution is $(1,0,0...0)$ max  and $(-1,0,0,...0)$ min. but for example $-e_2$ is also a max.
Can someone show me how to get the rigorous solution?
Here, $||\vec{x}||=\sqrt{x_1^2+x_2^2+\ldots+x_n^2}$.
 A: Observe that the function attains both positive and negative values on the given set, so that maximum and minimum must have $x_i\neq 0$ for every $i=1,\ldots , n$.
It is easy to verify that
$$
\frac{\partial f}{\partial x_k} = k \, \frac{f(x)}{x_k}
\qquad (x_k \neq 0),
$$
so that, using Lagrange multipliers, we obtain the conditions
$$
k \, \frac{f(x)}{x_k} = \lambda\, x_k,
\qquad k = 1, \ldots,n.
$$
Multiplying by $x_k$ and summing in $k$ we get
$$
f(x) \frac{n(n+1)}{2} = \lambda,
$$
so that our conditions, together with the fact that $f(x) \neq 0$ on max and min, gives
$$
x_k^2 = \frac{2k}{n(n+1)}, \qquad k = 1, \ldots, n,
$$
i.e.
$$
|x_k| = \sqrt{\frac{2k}{n(n+1)}}, \qquad k = 1, \ldots, n.
$$
Clearly, you get a min when the signs of the $x_k$ are choosen so that $f(x)$ is negative and a max when it is positive.
A: Per request by the OP, this is a solution using the AM-GM Inequality. I shall optimize
$$f(x):=\prod_{i=1}^n\,x_i^i\,,$$
where $x=(x_1,x_2,\ldots,x_n)\in\mathbb{R}^n$ satisfies
$$x_1^2+x_2^2+\ldots+x_n^2=1\,.$$
By the AM-GM Inequality,
$$\begin{align}
1&=\sum_{i=1}^n\,x_i^2=\sum_{i=1}^n\,i\,\left(\frac{x_i^2}{i}\right)
\\&\geq \left(\sum_{i=1}^n\,i\right)\left(\prod_{i=1}^n\,\left(\frac{x_i^2}{i}\right)^i\right)^{\frac{1}{\sum\limits_{i=1}^n\,i}}
=\frac{n(n+1)}{2}\,\frac{\left|\prod\limits_{i=1}^n\,x_i^i\right|^{\frac{4}{n(n+1)}}}{\left(\prod\limits_{i=1}^n\,i^i\right)^{\frac{2}{n(n+1)}}}\,.
\end{align}$$
This shows that
$$\left|\prod_{o=1}^n\,x_i^i\right|\leq \sqrt{\left(\frac{2}{n(n+1)}\right)^{\frac{n(n+1)}{2}}\,\prod_{i=1}^n\,i^i}\,.$$
The equality holds if and only if
$$|x_i|=\sqrt{\frac{2i}{n(n+1)}}\text{ for }i=1,2,\ldots,n.\tag{*}$$
By considering the signs, we conclude that the minimum value of $f(x)$ is
$$-\sqrt{\left(\frac{2}{n(n+1)}\right)^{\frac{n(n+1)}{2}}\,\prod_{i=1}^n\,i^i}\,,$$
which happens iff $x_1,x_2,\ldots,x_n$ satisfy (*), and an odd number of them are negative; the maximum value of $f(x)$ is
$$+\sqrt{\left(\frac{2}{n(n+1)}\right)^{\frac{n(n+1)}{2}}\,\prod_{i=1}^n\,i^i}\,,$$
which happens iff $x_1,x_2,\ldots,x_n$ satisfy (*), and an even number of them are negative.  For each positive integer $n$, there are precisely $2^{n-1}$ minimizing points, and $2^{n-1}$ maximizing points.
