Problem 1214

Let's rewrite the inequality:

$$\left(a_{n}-1\right)\left(a_{n+2}-1\right) \leqslant 1-a_{n+1}^{2}$$

By looking at the signs, we see that if $a_{k}>1$ for some $k \geqslant 3$, then necessarily (via $k=n+1$) either $a_{k-1}1>a_{k+1}$. By repeating this observation, we see that in the first case, we have $a_{k-1}, a_{k+2}1$, while in the second case, we have $a_{k-2}, a_{k+1}1$. Set $n=k-1$ in the first case, $n=k-2$ in the second; we then have:

$$\begin{aligned} & 1-a_{n+1}^{2} \geqslant\left(1-a_{n}\right)\left(1-a_{n+2}\right) \\ & 1-a_{n+2}^{2} \geqslant\left(1-a_{n+3}\right)\left(1-a_{n+1}\right) \end{aligned}$$

Or $\left(1-a_{n+3}\right)\left(1-a_{n+1}\right)=\frac{\left(1-a_{n+3}\right)\left(1-a_{n+1}^{2}\right)}{1+a_{n+1}} \geqslant \frac{\left(1-a_{n+3}\right)\left(1-a_{n}\right)\left(1-a_{n+2}\right)}{1+a_{n+1}}$.

Thus $\left(1+a_{n+1}\right)\left(1+a_{n+2}\right)\left(1-a_{n+2}\right) \geqslant\left(1-a_{n+3}\right)\left(1-a_{n}\right)\left(1-a_{n+2}\right)$.

But $1-a_{n+2}<0$ and $\left.1-a_{n}, 1-a_{n+3} \in\right] 0,1\left[\right.$, so $\left(1+a_{n+1}\right)\left(1+a_{n+2}\right)<1$, which is a contradiction.