Let {$a_n$} be a sequence satisfying $0<a_n<1$ and $4a_{n+1}(1-a_n) \ge 1$ $ \forall n \in \Bbb{N}$. Show that {$a_n$} converges to $\frac{1}{2}$.
My solution:
Using the Arithmetic Mean-Geometric Mean Inequality:
$\frac{a_{n+1}+(1-a_n)}{2} \ge \sqrt{a_{n+1}(1-a_{n})} > \frac{1}{2}$. Therefor $a_{n+1} > a_n$. Suppose $a_n \to a_0$ for some $a_0$. Then $4a_0(1-a_0) \ge 1$ which implies that $(2a_0-1)^2 \le 0$. Therefore $a_0 = \frac{1}{2}.$
Is my solution correct?
That's all fine, except the strict inequality. Note that the constant sequence $a_n = \frac{1}{2}$ satisfies all the conditions, but doesn't satisfy $$\frac{a_{n+1} + 1 - a_n}{2} \ge \sqrt{a_{n+1}(1 - a_n)} > \frac{1}{2}.$$ But this is OK, since the monotone convergence theorem doesn't require strict convergence to work anyway.