Find the extrema points of $ f\left(x\right)=\prod_{i=1}^{n}x_{i} $ Let $ f:\mathbb{R}^n \to \mathbb{R} $ given by
$ f\left(x\right)=\prod_{i=1}^{n}x_{i} $
I have to find the extrema points of $ f $ under the following constraint:
$ S=\left\{ \left(x_{1},...,x_{n}\right):\sum_{i=1}^{n}x_{i}^{p}=1\right\}  $
Where $ p $ is some positive even integer.
I tried to solve using Lagrange multipliers methood but it bacame too complicated so I doubt that thats the way to solve the question.
I'd appreaciate some help. Thanks in advance.
 A: I think Lagrange's multipliers work fine.
Namely, there is $\lambda$ such that $$L(x_{1},x_{2},...) = x_{1}*x_{2}*...*x_{n} - \lambda(x_{1}^{p} + x_{2}^{p} + ... x_{n}^{p}) $$
has all derivatives zero at the critical point.
Taking derivate with respect to $x_{1}$ one gets
$$x_{2}*x_{3}*...x_{n} -p*\lambda *x_{1}^{p-1} = 0$$
Multiplying both sides by $x_{1}$ gives
$$x_{1}*x_{2}*x_{3}*...x_{n} -p*\lambda *x_{1}^{p} = 0$$
Doing the same for each $k$ and then adding up leads to
$$nx_{1}*x_{2}*x_{3}*...x_{n} - p* \lambda(x_{1}^{p} + x_{2}^{p} + ... x_{n}^{p}) = 0$$
Which means, that at any extreme value one has
$$x_{1}*x_{2}*x_{3}*...x_{n} = p* \lambda *\frac{1}{n}$$
So, $$x_{2}*x_{3}*...x_{n} = p*\lambda  *\frac{1}{n * x_{1}}$$
Substituting this to the second equation from the top we finally get
$$ p*\lambda  *\frac{1}{n * x_{1}} = p*\lambda* x_{1}^{p-1}$$
and $$ x_{1}^{p} = \frac{1}{n}$$
Now, since $p$ is even $$x_{1} = +-\frac{1}{n^{\frac{1}{p}}}$$
A comment is in order, in the calculations above a few times we used the fact that
$x_{k}$ and $\lambda$ are not $0$ but this is obvious because at extreme points these values can not to be $0$ (otherwise max or min would be $0$ which is not the case). 
To finish, we notice that to form a maximum even $x_{k}$ have to come with sing $-$ and for minimum an odd number has to have sign $-$.
