Junior Balkan Mathematical Olympiad

In order to have $x^2 + ax + a^2 - 6 \neq 0$, 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 $\Delta = 3(8-a^2) \geq 0$, then
$$a^2 \leq 8 \Rightarrow a \leq 2\sqrt{2} \Rightarrow \frac{1}{a} \geq \frac{\sqrt{2}}{4}.$$
The given condition becomes
$$a^2 = 6 + \frac{6}{a} \geq 6 + \frac{6\sqrt{2}}{4} = 6 + \frac{3\sqrt{2}}{2} > 8, \text{ (contradiction).}$$