With $\vec{x}=(x_1,\ldots,x_n)$, find all the min and max of $\prod_{i=1}^{n} x_{i}^i$ given that $||\vec{x}||=1$

Question

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, $||\vec{x}||=\sqrt{x_1^2+x_2^2+\ldots+x_n^2}$.

Answer 1

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.

Answer 2

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.