Saudi Arabian Mathematical Competitions

We have $t^{2}+(k+2) t+k^{2}+k+1 \geq \frac{3}{4} k^{2}, \forall t \in \mathbb{R}$.
Indeed, this inequality is equivalent to
$$t^{2}+(k+2) t+\frac{1}{4} k^{2}+k+1 \geq 0$$
hence
$$t^{2}+(k+2) t+\left(\frac{1}{2} k+1\right)^{2} \geq 0$$
that is
$$\left(t+\frac{1}{2} k+1\right)^{2} \geq 0$$
We have equality if and only if $t=-\frac{1}{2} k-1$.
It follows that for any real numbers $x_{1}, x_{2}, \ldots, x_{n}$ we have
$$\prod_{k=1}^{n}\left(x_{k}^{2}+(k+2) x_{k}+k^{2}+k+1\right) \geq \prod_{k=1}^{n} \frac{3}{4} k^{2}=\left(\frac{3}{4}\right)^{n}(n!)^{2}$$
with equality if and only if $x_{k}=-\frac{1}{2} k-1, k=1,2, \ldots, n$, giving the unique solution to the problem.