Prove that the minimum value of all products of two values is less than $-\frac{1}{n}$. 
Let $x_1, x_2, \dots, x_n$ be $n$ real numbers, where $n \geq 3$ is an odd natural number. We know that:
$$x_1^2 + x_2^2 + \dots + x_n^2 = 1$$
and:
$$x_1^3 + x_2^3 + \dots + x_n^3 = 0$$
Prove that the minimum value of a product of type $x_ix_j$, with $0 < i < j < n +1$ is less than $\frac{-1}{n}$.

My thoughts: Let's consider two points $P$ and $Q$ in the $n$-dimensional topological space, one defined by $P(x_1, x_2, \dots, x_n)$ and the other by $Q(x_1^2, x_2^2, \dots, x_n^2)$. Obviously, from the first hypothesis, $P$ is located on the boundary of the $n - 1$-hypersphere. While, from the second condition, the scalar product $\vec{OP} \cdot \vec{OQ} = 0$, therefore $OP \perp OQ$ (perpendicularity defined in the $n$-space). From here, I want to find a $n$-geometrical condition for the collection of products, make their sum, and impose som condition on the smallest term.

Denote $a = \underset{1\le i\le n}{\min}{x_i}$ and $b = \underset{1\le i\le n}{\max}{x_i}$, then
$$\begin{align}
\sum_{i=1}^n (x_i^3-a^3)(x_i^3-b^3)\le0 &\Longleftrightarrow \sum_{i=1}^n x_i^6-\underbrace{\left( \sum_{i=1}^n x_i^3 \right)}_{=0}(a^3+b^3)+na^3b^3\le 0 \\
&\Longleftrightarrow a^3b^3\le -\frac{1}{n}\sum_{i=1}^n x_i^6  \tag{1} \\
\end{align}$$
Applying the Holder's inequality for $(p,q)= \left(3,\frac{3}{2} \right)$
$$\left(\sum_{i=1}^n \left(x_i^2\right)^{3} \right)^{1/3}\left(\sum_{i=1}^n 1^{3/2} \right)^{2/3} \ge \sum_{i=1}^n x_i^2 =1$$
$$\Longleftrightarrow \sum_{i=1}^n x_i^6 \ge \frac{1}{n^2} \tag{2}$$
From $(1),(2)$, we deduce that:
$$a^3b^3\le -\frac{1}{n^3} \Longleftrightarrow   \color{red}{ab\le -\frac{1}{n}}$$
The equality occurs if and only if $n=2k$ and for example $x_1=...=x_k = -\frac{1}{\sqrt{n}}$ and $x_{k+1}=...=x_{n} = \frac{1}{\sqrt{n}}$.
Hence, the minimum value of a product of type $x_ix_j$ is less than  $-\frac{1}{n}$ as $n$ is odd number.
Q.E.D
