Estonian Math Competitions

The inequality is equivalent to the inequality $na - n \le a^{n+1} - a$, or the inequality $n(a - 1) \le a(a - 1)(a^{n-1} + a^{n-2} + \dots + 1)$. If $a = 1$, then the inequality obviously holds. If $a < 1$ then $a - 1 < 0$, and dividing both sides of the inequality by $a - 1$ we get an equivalent inequality $n \ge a(a^{n-1} + a^{n-2} + \dots + 1)$, or $n \ge a^n + a^{n-1} + \dots + a$. Since $a < 1$, in the last sum all terms are less than 1, hence the sum does not exceed $n$.

If $n = 0$ then the inequality is $a \le a$, which obviously holds. Suppose that the inequality holds for $n = k$ and prove that it then holds for $n = k + 1$ as well. From the inequality $(k+1)a \le k + a^{k+1}$ we get the inequality $(k+2)a \le k + 1 + a^{k+2}$ by adding the inequality $a \le 1 + a^{k+2} - a^{k+1}$. The last inequality is equivalent to the inequality $(1-a)(1-a^{k+1}) \ge 0$, which obviously holds, because both factors are nonnegative.

If $a = 0$, then the inequality is $0 \le n$, which obviously holds. If $a > 0$, then AM-GM gives
$$\frac{1 + \dots + 1 + a^{n+1}}{n+1} \ge \sqrt[n+1]{1 \dots \cdot 1 \cdot a^{n+1}},$$
which is obviously equivalent with the original inequality.