Croatian Mathematical Competitions

For $a = 1$ the expression is equal to $-2$. We will prove that this is the minimal value, i.e. that for every $a \ge 0$ we have $a^3 - a^2 - 2\sqrt{a} \ge -2$.
We have
$$a^3 - a^2 - 2\sqrt{a} + 2 = a^2(a-1) - 2(\sqrt{a}-1) = (\sqrt{a}-1)(a^2(\sqrt{a}+1) - 2).$$
If $a \ge 1$, then $a^2(\sqrt{a}+1) - 2 \ge 2 - 2 = 0$ and $\sqrt{a}-1 \ge 0$, so our statement is true.
If $0 \le a < 1$, then $a^2(\sqrt{a}+1) - 2 < 2 - 2 = 0$ and $\sqrt{a}-1 < 0$, so our statement is true again.