IMO Hk TST

Let us denote $b_n = a_n$ for all $n$. The given inequality is:
$$(a_{n+1})^2 + a_n a_{n+2} \le a_n + a_{n+2}$$
for all $n \ge 1$.

We can rearrange the inequality:
$$(a_{n+1})^2 + a_n a_{n+2} - a_n - a_{n+2} \le 0$$
$$(a_{n+1})^2 - a_n + a_n a_{n+2} - a_{n+2} \le 0$$
$$(a_{n+1})^2 - a_n + (a_n - 1)a_{n+2} \le 0$$

Let us try to bound $a_n$ from above. Suppose $a_n > 1$ for some $n$. Since all $a_n > 0$, we can divide both sides by $a_n$ (if needed), but let's try to use the structure.

Suppose $a_n \le 1$ for all $n$. Then $a_{2022} \le 1$ as desired. Suppose not, and let $k$ be the smallest index such that $a_k > 1$.

Since $a_1 > 0$, and $a_1 \le 1$ (otherwise $k=1$), so $a_1 \le 1$, $a_2 \le 1$, ..., $a_{k-1} \le 1$, $a_k > 1$.

Now, consider the inequality for $n = k-1$:
$$(a_k)^2 + a_{k-1} a_{k+1} \le a_{k-1} + a_{k+1}$$
$$(a_k)^2 \le a_{k+1} (1 - a_{k-1}) + a_{k-1}$$
But $a_{k-1} \le 1$, so $1 - a_{k-1} \ge 0$, and $a_{k+1} > 0$.
So $a_{k+1} (1 - a_{k-1}) + a_{k-1} \le a_{k+1} + a_{k-1}$.
But $(a_k)^2 \le a_{k+1} (1 - a_{k-1}) + a_{k-1} \le a_{k+1} + a_{k-1}$.
But $a_k > 1$, so $(a_k)^2 > 1$.
Therefore, $a_{k+1} (1 - a_{k-1}) + a_{k-1} > 1$.

But $a_{k-1} \le 1$, so $a_{k+1} (1 - a_{k-1}) > 1 - a_{k-1} \ge 0$.
So $a_{k+1} > 1$ unless $a_{k-1} = 1$.
If $a_{k-1} = 1$, then $a_{k+1} (1 - a_{k-1}) = 0$, so $(a_k)^2 \le 1$.
But $a_k > 1$, so $(a_k)^2 > 1$, contradiction.

Therefore, $a_{k-1} < 1$, so $1 - a_{k-1} > 0$, and $a_{k+1} > \frac{1 - a_{k-1}}{1 - a_{k-1}} = 1$.
So $a_{k+1} > 1$.

Thus, if $a_k > 1$ for some $k$, then $a_{k+1} > 1$ as well. But then, by induction, $a_n > 1$ for all $n \ge k$.

Now, consider the original inequality for $n \ge k$:
$$(a_{n+1})^2 + a_n a_{n+2} \le a_n + a_{n+2}$$
But $a_n > 1$, $a_{n+2} > 1$, so $a_n + a_{n+2} < (a_n)(a_{n+2})$ (since $a_n, a_{n+2} > 1$).
So $a_n a_{n+2} > a_n + a_{n+2}$.
Therefore,
$$(a_{n+1})^2 + a_n a_{n+2} > (a_{n+1})^2 + a_n + a_{n+2}$$
But the inequality says $(a_{n+1})^2 + a_n a_{n+2} \le a_n + a_{n+2}$, so $(a_{n+1})^2 \le 0$, which is impossible since $a_{n+1} > 0$.

Therefore, our assumption that $a_k > 1$ for some $k$ leads to a contradiction. Thus, $a_n \le 1$ for all $n$, in particular $a_{2022} \le 1$.