Let \((a_n)_{n\geq 1}\) be a sequence of positive real numbers with the property that
\[(a_{n+1})^2 + a_na_{n+2} \leq a_n + a_{n+2}\]
for all positive integers \(n\). Show that \(a_{2022}\leq 1\).We begin by observing that \(\left(a_{n+1}\right)^{2}-1 \leqslant a_{n}+a_{n+2}-a_{n} a_{n+2}-1\), which is equivalent to
\[\left(a_{n+1}\right)^{2}-1 \leqslant\left(1-a_{n}\right)\left(a_{n+2}-1\right)\]
Suppose now that there exists a positive integer \(n\) such that \(a_{n+1}>1\) and \(a_{n+2}>1\). Since \(\left(a_{n+1}\right)^{2}-1 \leqslant\left(1-a_{n}\right)\left(a_{n+2}-1\right)\), we deduce that \(0<1-a_{n}<1<1+a_{n+2}\), thus \(\left(a_{n+1}\right)^{2}-1<\left(a_{n+2}+1\right)\left(a_{n+2}-1\right)=\left(a_{n+2}\right)^{2}-1\). On the other hand, \(\left(a_{n+2}\right)^{2}-1 \leqslant\left(1-a_{n+3}\right)\left(a_{n+1}-1\right)<\left(1+a_{n+1}\right)\left(a_{n+1}-1\right)=\left(a_{n+1}\right)^{2}-1\), a contradiction. We have shown that we cannot have two consecutive terms, except maybe \(a_{1}\) and \(a_{2}\), strictly greater than \(1\). Finally, suppose \(a_{2022}>1\). This implies that \(a_{2021} \leqslant 1\) and \(a_{2023} \leqslant 1\). Therefore \(0< \left(a_{2022}\right)^{2}-1 \leqslant\left(1-a_{2021}\right)\left(a_{2023}-1\right) \leqslant 0\), a contradiction. We conclude that \(a_{2022} \leqslant 1\).