Saudi Arabia Mathematical Competitions

We have $t^2+(k+2)t+k^2+k+1\ge \frac{3}{4}{k}^2,\forall t\in \mathbb{R}$. Indeed, this inequality is equivalent to $$t^2+(k+2)t+\frac{1}{4}{k}^2+k+1\ge 0$$ hence $$t^2+(k+2)t+{\left(\frac{1}{2}k+1\right)}^2\ge 0$$ that is $${\left(t+\frac{1}{2}k+1\right)}^2\ge 0$$ We have equality if and only if $t=-\frac{1}{2}k-1$. It follows that for any real numbers $x_1,x_2,\dots ,x_n$ we have $$\prod _{k=1}^n\left(x_k^2+(k+2)x_k+k^2+k+1\right)\ge \prod _{k=1}^n\frac{3}{4}{k}^2={\left(\frac{3}{4}\right)}^n(n!{)}^2$$ with equality if and only if $x_k=-\frac{1}{2}k-1,k=1,2,\dots ,n$, giving the unique solution to the problem.