Olimpiade Italiana di Matematica

Solution:

We shall prove that the inequality
- holds for every choice $x_{1}, \ldots, x_{n}$ of positive real numbers if and only if $n \geq 4$;
- holds for every choice $x_{1}, \ldots, x_{n}$ of real numbers if and only if $n=4$.

We split the proof into several steps.

Step 1. We show that for $n=3$ the inequality is false, even for triples of positive real numbers. Indeed, setting $x_{1}=x_{2}=x_{3}=1$, the left-hand side equals $9$ while the right-hand side equals $12$.

Step 2. For $n=4$ the inequality holds for every choice of real numbers $x_{1}, x_{2}, x_{3}, x_{4}$ (positive, negative, or zero).
Indeed, expanding the square and moving everything to the left-hand side we obtain the inequality
$$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}-2 x_{1} x_{2}-2 x_{2} x_{3}-2 x_{3} x_{4}-2 x_{4} x_{1}+2 x_{1} x_{3}+2 x_{2} x_{4} \geq 0$$
that is
$$\left(x_{1}-x_{2}+x_{3}-x_{4}\right)^{2} \geq 0$$
which is trivially always true.

Step 3. For every $n \geq 5$ there exists at least one choice $x_{1}, \ldots, x_{n}$ of real numbers (not all positive) for which the inequality is false.
Indeed, setting $x_{1}=x_{2}=1, x_{3}=0, x_{4}=-2, x_{5}=\ldots=x_{n}=0$ the left-hand side vanishes, while the right-hand side equals $4$.

Step 4. For every $n \geq 4$ the inequality holds for every choice $x_{1}, \ldots, x_{n}$ of positive real numbers.
We prove this fact by induction on $n$. Suppose the claim holds for $n \geq 4$ and let us prove it for $n+1$. Let $x_{1}, \ldots, x_{n+1}$ be positive real numbers. Note that the inequality does not change if we cyclically permute the numbers $x_{1}, \ldots, x_{n+1}$, and therefore we may assume that $x_{1}$ is a number greater than or equal to all the others. From the inequality for $n$ terms we obtain
$$\begin{aligned} \left(x_{1}+\cdots+x_{n-1}+\left(x_{n}+x_{n+1}\right)\right)^{2} \geq & 4\left(x_{1} x_{2}+\cdots+x_{n-2} x_{n-1}+\right. \\ & \left.+x_{n-1}\left(x_{n}+x_{n+1}\right)+\left(x_{n}+x_{n+1}\right) x_{1}\right) \\ = & 4\left(x_{1} x_{2}+\cdots+x_{n} x_{n+1}+x_{n+1} x_{1}\right)+4 y \end{aligned}$$
where $y=x_{n-1} x_{n+1}+x_{n}\left(x_{1}-x_{n+1}\right)>0$ by our assumption on $x_{1}$. The inequality is proved.