# Global maxima of a function subject to a constraint.

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.

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.