11th Junior Balkan Mathematical Olympiad

In order to have $x^2+ax+a^2-6\ne 0$, for all $x\in \mathbb{R}$, it is enough the discriminant $\Delta =3(8-a^2)$ to be less than $0$, that is $\Delta =3(8-a^2)<0$.

In fact, if we suppose that $\Delta =3(8-a^2)\ge 0$, then $$a^2\le 8\Rightarrow a\le 2\sqrt{2}\Rightarrow \frac{1}{a}\ge \frac{\sqrt{2}}{4}.$$ The given condition becomes $$a^2=6+\frac{6}{a}\ge 6+\frac{6\sqrt{2}}{4}=6+\frac{3\sqrt{2}}{2}>8,\text{ (contradiction).}$$