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?
 A: 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 $1/2$.
A: Claim: The maximum occurs when $ \{ x_i \} \in \{ \alpha , \beta \}$.
Proof: Consider any $ x_i, x_j, x_k$.
WLOG $x_i x_j x_k >  0 $.
Let $ a = x_i + x_j + x_k, b = x_ix_j + x_jx_k+x_kx_i$.
Consider the cubic equation $ x^3  - ax^2 + bx - c = 0 $.
Then, since we want real roots, we know that $c$ is maximized at a turning point of the cubic. This means we can replace $ (x_i, x_j, x_k)$ with a corresponding $(x_i' = x_j' , x_k')$ while increasing $ \prod x_i$.
Hence, at the maximum, out of any 3 variables, there exists (at least) 2 of them which are equal.
Thus, $ \{ x_i \} \in \{ \alpha , \beta \}$.
Corollary: By setting $ x_1 = \ldots = x_i = \alpha, x_{i+1} = \ldots = x_6 = \beta$, solving the simultaneous equations $ i \alpha + (6-i) \beta = 0, i \alpha^2 + (6-i) \beta^2 = 0 $, and calculating $ \alpha^i \beta^{6-i}$ for each $i$, we conclude that the maximum happens when $ i = 4, \alpha = 1 / \sqrt{2}, \beta = \sqrt{2}, \prod x_i = 1/2$.
Note: The same holds for finding the minimum (after establishing that it is <0), which happens when $ i = 3, \alpha = 1, \beta = -1, \prod x_i = - 1 $.
