Find the max of $x_1 x_2 \cdots x_m$ when $x_1^2+x_2^2+ \cdots + x_m^2=1$ 
Find the max of a function $$f(x_1, x_2,...,x_m)=x_1 x_2 \cdots x_m$$
  when $x_1^2+x_2^2+ \cdots +x_m^2=1$

I am looking for unique solutions. So far I've tried two different ways.
One of them was with Lagrange multiplier but there seem to be quite too many variables and I couldn't express $x_1,...,x_m$ separately. 
The second one solution is with the Inequality of arithmetic and geometric means.
$$\frac{x_1^2+...+x_m^2}{m}\ge \sqrt[m]{x_1^2 \cdots x_m^2} \implies \frac{1}{m} \ge \sqrt[m]{x_1^2 \cdots x_m^2}=(x_1 \cdots x_m)^{\frac{2}{m}} \implies $$ $$x_1 \cdots x_m \le \frac{1}{m^{\frac{m}{2}}}$$ So we can assume that $$\max f(x_1, x_2,...,x_m)=\frac{1}{m^{\frac{m}{2}}} \quad \forall x_i=\frac{1}{\sqrt m}, \quad i=1,2,...,m.$$
I'd love to know if somebody solved this exercise with Langrange multiplier or in other way. Thanks in advance!
 A: $$x_1x_2...x_m\neq\sqrt{x_1^2x_2^2...x_m^2}$$
The right way it's the following:
By AM-GM 
$$x_1x_2...x_m\leq\sqrt{x_1^2x_2^2...x_m^2}\leq\sqrt{\left(\frac{x_1^2+x_2^2+...+x_m^2}{m}\right)^m}=\frac{1}{m^{\frac{m}{2}}}$$
The equality occurs for $x_1=x_2=...=x_m=\frac{1}{\sqrt{m}},$ which says that we got a maximal value.
A: Let $g(x) = ||x||_2^2 - 1 = 0$ be the constraint.  $\nabla g(x) = 2x$, $$\nabla f(x)= (f(x)/x_1,\dots,f(x)/x_m) = \lambda \nabla g(x) = 2\lambda x.$$
$$\frac{x_1\cdots x_m}{x_i} = 2 \lambda x_i \\ x_1\cdots x_m = 2 \lambda x_i^2$$
Observe the the LHS is symmetric by interchange of indices $i$ and $j$ in $\{1,\dots,m\}$, so $x_i = \pm x_j$.
Sum the above inequality over $i$ to get $m x_1 \cdots x_m = 2 \lambda$.  This should give $|x_i| = \sqrt[m]{\dfrac{2|\lambda|}{m}}$.
Substitute this back to $g(x) = 0$: $m \cdot \sqrt[m]{\dfrac{4\lambda^2}{m^2}} = 1,$ so $\dfrac{2|\lambda|}{m} = \pm \dfrac{1}{\sqrt{m^m}},$ giving a critical point
$x^* = \left(\dfrac{1}{\sqrt{m}},\dots,\dfrac{1}{\sqrt{m}}\right)$
when $\lambda^* = \dfrac{m^{1-m/2}}{2}$, which is the same as the one found by AM-GM inequality in the question body.
Evaluating the objective at $x^*$, we find that
$f(x^*) = x_1^* x_2^* \dots x_m^* = \dfrac{1}{\sqrt{m^m}}$.
To determine whether this is a local maximum, define the Lagrangian
$$\begin{aligned}{\mathcal {L}}(x,\lambda) &= f(x)-\lambda g(x)\\
\nabla{\cal L}(x,\lambda) &= \left({\frac {\partial {\mathcal {L}}}{\partial x}},{\frac {\partial {\mathcal {L}}}{\partial \lambda }}\right) = \left(\frac{x_1\cdots x_m}{x_1} - 2\lambda x_1, \dots, \frac{x_1\cdots x_m}{x_m} - 2\lambda x_m, -\sum_{i=1}^m x_i^2 \right)
\end{aligned}$$
We compute the Hessian matrix of ${\cal L}$ at $x^*$.  For all $i,j \in \{1,\dots, m\}$ with $i \ne j$,
\begin{alignat}{2}
\frac{\partial^2{\mathcal {L}}}{\partial x_i^2} &= -2\lambda \quad &
\frac{\partial^2{\mathcal {L}}}{\partial x_i \partial x_j} &= \frac{x_1\cdots x_m}{x_i x_j} \\
\frac{\partial^2{\mathcal {L}}}{\partial \lambda^2} &= 0 &
\frac{\partial^2{\mathcal {L}}}{\partial \lambda \partial x_i} &= -2 x_i
\end{alignat}
This gives a symmetric matrix
\begin{bmatrix}
-2\lambda & \dfrac{x_1\cdots x_m}{x_1 x_2} & \dots & \dfrac{x_1\cdots x_m}{x_1 x_m} & -2 x_1 \\
\dfrac{x_1\cdots x_m}{x_1 x_2} & -2\lambda & \dots & \dfrac{x_1\cdots x_m}{x_2 x_m} & -2 x_2 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
\dfrac{x_1\cdots x_m}{x_1 x_m} & \dfrac{x_1\cdots x_m}{x_2 x_m} & \dots & -2\lambda & -2 x_m \\
-2x_1 & -2x_2 & \dots & -2x_m & 0
\end{bmatrix}
Substituting $(x^*,\lambda^*) = (m^{-1/2},\dots,m^{-1/2},m^{1-m/2}/2)$, we have
$$H = \begin{bmatrix}
-m^{1-m/2} & m^{1-m/2} & \dots & m^{1-m/2} & -2 m^{-1/2} \\
m^{1-m/2} & -m^{1-m/2} & \dots & m^{1-m/2} & -2 m^{-1/2} \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
m^{1-m/2} & m^{1-m/2} & \dots & -m^{1-m/2} & -2 m^{-1/2} \\
-2 m^{-1/2} & -2 m^{-1/2} & \dots & -2 m^{-1/2} & 0
\end{bmatrix}$$
For $k = 1,\dots,m$, let $M_k$ be the $k$-th leading principal minor of $-H$.  Then the $M_k$'s are circulant-matrices whose associated polynomial is $f(x) = C \left( 1-\sum\limits_{\ell=1}^{k-1} x^\ell \right)$, where $C = m^{1-m/2} \in (0,1)$ for $m \ge 3$.
Let $\omega = e^{2\pi i /k}$.  A simple geometric sum would give 
$$
\sum_{\ell=0}^{k-1} (\omega^j)^\ell =
\begin{cases}
0 &\text{ if } j = 1,\dots,k-1 \\
k &\text{ if } j = 0
\end{cases}
$$
Then $\det(M_k) = \prod\limits_{j = 0}^{k-1} f(\omega^j) = C^k \cdot [-(k-2)] \cdot 2^{k-1} < 0$ for all $k > 3$.
A: Either the function and the constraint are symmetrical wrt $x_1=x_2= \cdots=x_m$.   
The function $f(x_1,\cdots,x_m)=x_1 x_2 \cdots x_m=s$ is one of the sheets of a  2-sheets hyperboloid,
of revolution around the axis above, with minimum (apex) at $x_1= \cdots =x_m=s^{1/m}$.
The constraint is a sphere.
Thus .. 
A: I would do it another way.  The symmetry suggests that the maximum will occur when all the coordinates are equal, so I would try to show that if two coordinates say $x_i$ and $x_j$ are different, then we can increase the objective function by replacing both of them by some value $c$ and leaving the other coordinates the same.  Since we have to maintain the constraint, this comes down to showing that the maximum of $g(x,y)=xy$ subject to $x^2+y^2=k$ is attained when $x=y$  This is easy with Lagrange multipliers.  
Once we know that $x_1=x_2=\cdots=x_m$ we solve for the common value.
