Maximize $\prod x_i$ without Lagrange multipliers 
Suppose that $\sum_{i=1}^{6} x_i=0$ and $\sum_{i=1}^{6} x_i^2=6$. We wish to maximize $\prod x_i$.

One can show via Lagrange multipliers that at an extremal point, either $x_i=x_j$ or $x_i x_j=-1$, which leads to the maximum of $1/2$ when 4 of the variables are $1/\sqrt 2$ and the other two are $-\sqrt 2$.
However, given the niceness of the functions involved, it feels like there should be a more elementary approach to showing that $1/2$ is an upper bound.
Question: Is there a simpler alternative approach?

Let $p$ denote the maximum of $\prod x_i$. Clearly, $p > 0$.
We only need to consider two cases:
Case 1: $x_1, x_2, x_3, x_4 > 0$ and $x_5, x_6 < 0$
Denote $A = x_1 + x_2 + x_3 + x_4 > 0$.
Then $x_5 + x_6 = -A$.
Clearly, we have $x_5^2 + x_6^2 \ge (x_5 + x_6)^2/2 = A^2/2$.
Also, $x_1^2 + x_2^2 + x_3^2 + x_4^2 \ge (x_1 + x_2 + x_3 + x_4)^2/4 = A^2/4$.
Thus, we have  $6 = \sum_{i=1}^6 x_i^2 \ge A^2/2 + A^2/4$
which results in $A \le 2\sqrt 2$.
Using AM-GM, we have
$$x_1x_2x_3x_4 \le \frac{(x_1 + x_2 + x_3 + x_4)^4}{4^4} = A^4/4^4.$$
Also, we have
$$x_5x_6 \le \frac{(x_5 + x_6)^2}{4} = A^2/4.$$
Thus, we have
$$\prod_{i=1}^6 x_i
\le \frac{A^4}{4^4}\, \frac{A^2}{4} = \frac{A^6}{4^5}
\le \frac{1}{2}$$
with equality if $x_1 = x_2 = x_3 = x_4 = 1/\sqrt 2$
and $x_5 = x_6 = -\sqrt 2$.
Case 2: $x_1, x_2 > 0$ and $x_3, x_4, x_5, x_6 < 0$
Similar.
Thus, the maximum of $\sum_{i=1}^6 x_i$ is $1/2$.
