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I am trying to prove that the global maxima of $f(x_1,x_2,...,x_n)=(x_1x_2···x_n)^2$, subject to $\lVert(x_1,x_2,...,x_n)\rVert_2=r$ is $(r^2/n)^n$

I know I have to find the critical points of the Lagrange function, that is $$F(x_1,x_2,...,x_n,\lambda)=(x_1x_2···x_n)^2+\lambda(\sqrt{\sum_{i=1}^nx_i^2}-r)$$ In order to do that, I've built the system in which all partial derivatives are null, so I have:

$$ D_1F(x_1,x_2,...,x_n,\lambda)=2(x_1x_2···x_n)(x_2x_3···x_n)+\frac{\lambda x_1}{\sqrt{\sum_{i=1}^nx_i^2}}=0 $$ $$ ... $$ $$ D_jF(x_1,x_2,...,x_n,\lambda)=2(x_1x_2···x_n)(x_1x_2···x_{j-1}x_{j+1}···x_n)+\frac{\lambda x_j}{\sqrt{\sum_{i=1}^nx_i^2}}=0 $$ $$ ... $$ $$ D_nF(x_1,x_2,...,x_n,\lambda)=2(x_1x_2···x_n)(x_1x_2···x_{n-1})+\frac{\lambda x_n}{\sqrt{\sum_{i=1}^nx_i^2}}=0 $$ $$ D_{n+1}F(x_1,x_2,...,x_n,\lambda)=\sqrt{\sum_{i=1}^n}x_i^2-r=0 $$

I think, because I have tried to solve reduced forms of this system in a mathematical software, that the solutions are like the following: $$ (x_1,x_2,...,x_n)=(\pm\frac{r}{\sqrt{n}},\pm\frac{r}{\sqrt{n}},...\pm\frac{r}{\sqrt{n}}) $$ where all the $x_i$ take the positive and negative sign, so there are $2^n$ solutions.

However, I didn't manage to solve the system. I have tried to replace the summation with $r$ in all equations, to sum all the equations, to reduce dividing by $x_1···x_n$... but I didn't get to any interesting end.

Could you help me, please, giving me any hint about how can I solve this system?

Thank you very much.

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You can argue that since the function you want to maximize is convex and since all the $x_i$ appear symmetrically in the objective function as well as in the ristriction(s), that there is a solution in which all of them can be equal. So by guessing that $x_1=x_2=\ldots=x_n := x$ you can solve the system with one unknown (the $x$) and then it is straightforward to show that the global maximum is equal to $(r^2/n)^n$. Guessing a property of the solution is common in nonlinear programming.

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While I agree with @Stefanos that you must guess for symmetry in such problems (after all, why shouldn't it be symmetric? If one variable is more than another, what is pushing it to be?), I feel that you should approach such problems with as much simplification of the problem as possible. As an example, you should immediately get rid of the norm and replace it with sum of squares.

\begin{align} \min_x \quad & (x_1\cdot x_2 \cdots x_n)^2 \\ s.t. \quad & \sum _{i=1}^n x_i^2=r^2\\ L(x,\lambda)&= (x_1\cdot x_2 \cdots x_n)^2 + \lambda(\sum _{i=1}^n x_i^2=r^2)\\ \nabla_x L(x,\lambda)&= 2\begin{bmatrix} x_1(x_2 \cdots x_n )^2\\x_2(x_1x_3\cdots x_n )^2\\\vdots\\x_n(x_1x_2\cdots x_{n-1}) \end{bmatrix}+2\lambda \begin{bmatrix}x_1\\x_2\\ \vdots \\x_n \end{bmatrix}=0\\ &= \begin{bmatrix} x_1((x_2 \cdots x_n )^2+\lambda)\\x_2((x_1x_3\cdots x_n )^2+\lambda)\\\vdots\\x_n((x_1x_2\cdots x_{n-1})+\lambda) \end{bmatrix}\\ \end{align}

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  • $\begingroup$ Thank you, @Nitish, both of you have help me to understand what I was doing! :) $\endgroup$ – Alejandro Feb 19 '14 at 1:32

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