sign bash is usually a solution but never the official solution... what happened here

Solved with megarnie. Define $b_n=a_n-1$. Then \[-b_nb_{n+2}\ge b_{n+1}(b_{n+1}+2).\]We will show that $b_{2022}$ is nonpositive, so assume FTSOC it is positive. $n=2021$ gives $b_n$ and $b_{n+2}$ has opposite signs, then doing this on the one that is positive gives that there is a substring of the sequence that goes $-++-$. Let this substring be $-w$, $x$, $y$, $-z$. Then \[y\ge wy\ge x(x+2)\ge 2x\]\[x\ge xz\ge y(y+2)\ge 2y\]so $x=y=0$, contradiction. $\blacksquare$

Rewrite the equation as $a_{n+1}^2+(a_n-1)(a_{n+2}-1)\leq 1.$ If $a_1=a_2=a_3=1,$ then $a_{2023}=1$. Assume that not all of $a,b,c$ are equal to $1$ then. Claim: One of $a_1,a_2,a_3$ must be $<1$. Proof: Suppose not, then $a_1,a_2,a_3\geq 1$ with at least one strict inequality. Thus $a_2^2+(a_1-1)(a_3-1)>1,$ contradiction. Claim: For all $n$, we have $a_{n+1}^2<a_n.$ Proof: This is rearranging the above equation. We say \begin{align*} & a_{n+1}^2+(a_n-1)(a_{n+2}-1)\leq 1 \\ &\iff (a_n-1)(a_{n+2}-1)\leq 1-a_{n+1}^2 \\ &\iff a_{n+2}-1 \leq \frac{1-a_{n+1}^2}{a_n-1} \\ &\iff a_{n+2}\leq \frac{a_n-a_{n+1}^2}{a_n-1}. \end{align*}But $a_{n+2}>0$ so $0<a_n-a_{n+1}^2$ as desired. Putting the two claims together implies the result.

Notice that the inequality from the problem statement is equivalent to $$a_{n+1}^2-1\leqslant(a_n-1)(1-a_{n+2})\text{.}$$Let's consider sequences of symbols $<$ and $>$, where we assign the symbol $>$ to each $a_i$ iff $a_i > 1$, and we assign the symbol $<$ to each $a_i$ iff $a_i \leqslant 1$. There are 8 sequences of three symbol of them: $>>>$, $>><$, $><>$, $><<$, $<>>$, $<><$, $<<>$ and $<<<$. Array $>>>$ is not possible because $$0<a_{n+1}^2-1\leqslant(a_n-1)(1-a_{n+2})<0\text{.}$$Array $<><$ is not possible because $$0<a_{n+1}^2-1\leqslant(a_n-1)(1-a_{n+2})\leqslant 0\text{.}$$ Let's consider four consecutive elements of the sequence, $a_i$, $a_{i+1}$, $a_{i+2}$, and $a_{i+3}$. Let $a_i$ be denoted by $<$, and let $a_{i+1}$ and $a_{i+2}$ be denoted by $>$. Then $a_{i+3}$ must be denoted by $<$ (since otherwise $a_{i+1}$, $a_{i+2}$, and $a_{i+3}$ would form the sequence $>>>$, which we have already proved that does not exist in the sequence). However, it turns out that this is not possible: $$a_{i+1}^2-1\leqslant(a_i-1)(1-a_{i+2})\Leftrightarrow a_{i+1}^2-a_{i+2}\leqslant a_i-a_ia_{i+2}=a_i(1-a_{i+2})<0\Leftrightarrow a_{i+1}^2<a_{i+2}\text{,}$$analogously we get $a_{i+2}^2<a_{i+1}$ and from these two we get $a_{i+1}^4<a_{i+2}^2<a_{i+1}$, which is contradiction because $a_{i+1}>1$. Therefore, $<>>$ can't exist in an array because we can't extend it. We have five arrays left: $>><$, $><>$, $><<$, $<<>$ and $<<<$. Also note that strings $<<>$ and $><>$ can't exist in an array because we can't extend it (of the remaining five strings, there are none that begin with <>). We have three arrays left: $>><$, $><<$ and $<<<$. 1st case: array $a$ begin with $>><$, then $a_2$ is $>$, $a_3$ is $<$, $a_4$ is $<$, then every member of $a_n$ for $n\geqslant 5$ is $<$, so $\boxed{a_{2023}\leqslant 1}$. 2nd case: array $a$ begin with $><<$, then every member of $a_n$ for $n\geqslant 2$ is $<$, so $\boxed{a_{2023}\leqslant 1}$. 3rd case: array $a$ begin with $<<<$, then every member of $a_n$ is $<$, so $\boxed{a_{2023}\leqslant 1}$.

Slightly hard for A1?

Plug $b_i=1-a_i$ so $b_i\in (-\infty;1).$ We claim that $b_n>0$ for every $n>2,$ which suffices. Rewrite the inequality as $b_{n+1}(2-b_{n+1})\geq b_nb_{n+2}$ $(\bigstar).$ FTSOC $b_m<0$ for some $m>2.$ Due to the $(\bigstar)$ numbers $b_{m-1},$ $b_{m+1}$ have different signs; thus we can find $k$ such that $b_k>0,$ $b_{k+1}<0,$ $b_{k+2}<0,$ $b_{k+3}>0.$ By $(\bigstar)$ $$b_{k+1}b_{k+2}\geq b_kb_{k+2}\cdot b_{k+1}b_{k+3}\geq b_{k+1}(2-b_{k+1})\cdot b_{k+2}(2-b_{k+2})\implies 1\geq (2-b_{k+1})(2-b_{k+2}),$$which however fails. The conclusion follows.

Let $b_n=a_n+1$ then we have \[b_{n+1}^2+2b_{n+1}+2+b_nb_{n+2}\ge 0\]and all $b_n\ge -1$. If $b_{n+1}>0$ then $b_nb_{n+2}<0$ so if $b_n>0$ then for some $\varepsilon\in \{-1,1\}$, $b_{n+\varepsilon}>0$ and $b_{n-\varepsilon}<0$ and applying the same thing on $n+\varepsilon$ gives $b_{n+2\varepsilon}<0$. Now, \[b_{n+\varepsilon}^2+2b_{n+\varepsilon}+2\ge (-b_{n+2\varepsilon})(b_n)\ge b_n\]and similarly $b_n^2+2b_n+2\ge b_{n+\varepsilon}$ and summing the two equations gives a contradiction.

If $a_{2022} \leq a_{2021}$ then $a_{2022}^2 + a_{2020}a_{2022} \leq a_{2021}^2 + a_{2020}a_{2022} \leq a_{2020}+a_{2022}$, finishing. If $a_{2022} \leq a_{2023}$ then $a_{2022}^2 + a_{2022}a_{2024} \leq a_{2023}^2 + a_{2022}a_{2024} \leq a_{2022}+a_{2024}$, finishing. Otherwise, the same two inequalities yield $a_{2021}, a_{2023} \leq 1$. Now $a_{2022}^2 \leq 1 - (1-a_{2021})(1-a_{2023}) \leq 1$, finishing.

We uploaded our solution https://calimath.org/pdf/ISL2022-A1.pdf on youtube https://youtu.be/LCRnjfFuJ9Q.

GrantStar wrote: Rewrite the equation as $a_{n+1}^2+(a_n-1)(a_{n+2}-1)\leq 1.$ If $a_1=a_2=a_3=1,$ then $a_{2023}=1$. Assume that not all of $a,b,c$ are equal to $1$ then. Claim: One of $a_1,a_2,a_3$ must be $<1$. Proof: Suppose not, then $a_1,a_2,a_3\geq 1$ with at least one strict inequality. Thus $a_2^2+(a_1-1)(a_3-1)>1,$ contradiction. Claim: For all $n$, we have $a_{n+1}^2<a_n.$ Proof: This is rearranging the above equation. We say \begin{align*} & a_{n+1}^2+(a_n-1)(a_{n+2}-1)\leq 1 \\ &\iff (a_n-1)(a_{n+2}-1)\leq 1-a_{n+1}^2 \\ &\iff a_{n+2}-1 \leq \frac{1-a_{n+1}^2}{a_n-1} \\ &\iff a_{n+2}\leq \frac{a_n-a_{n+1}^2}{a_n-1}. \end{align*}But $a_{n+2}>0$ so $0<a_n-a_{n+1}^2$ as desired. Putting the two claims together implies the result. I might be wrong but I think your second claim might not be true. You seem to divide by $(a_n-1)$ in your proof while assuming that $a_n-1>0$. However it doesn't manage to show that $a_{n+1}^2<a_n$ for values of $a_n<1$

The ISL journey begins: Claim: For each $n \geq 2$, one of $a_n$, or $a_{n+1}$, is less or equal to $1$. Proof: $$a_n^2 +a_{n-1}a_{n+1} \leq a_{n-1}+a_{n+1}$$$$a_{n+1}^2+a_{n}a_{n+2} \leq a_n + a_{n+2} $$$$a_{n+1}^2+a_{n}a_{n+2}+a_n^2 +a_{n-1}a_{n+1} \leq a_n + a_{n+2}+ a_{n-1}+a_{n+1}$$$$(a_{n+1}-1)(a_{n+1}+a_{n-1})+ (a_{n}-1)(a_{n+2}+a_{n}) \leq 0$$So one of $a_{n}$, $a_{n+1}$ is less than $0$. $\square$ Now define a sequence $b_i= 1- a_i$. Suppose for the sake of contradiction $b_{2022} < 0$, then $b_{2021}, b_{2023} \geq 0$. Then: $$(1-b_{2022})^2+ (1-b_{2021})(1-b_{2023}) \leq 1-b_{2021}+1 - b_{2023}$$$$b_{2022}^2+b_{2023}b_{2021} \leq 2b_{2022}$$But the left hand side is positive and the right hand side is negative, contradiction $\blacksquare$

The claim is that the sequence $\{a_n\}$ must eventually have terms at most $1$. Rearrange the equation to read \[(1-a_{n+1})^2 \geq (1-a_n)(1-a_{n+2}).\]Claim. If $a_n > 1$, then one of $a_{n-1}$ or $a_{n+1}$ is at most $1$. Proof. Suppose otherwise. Then $(1-a_{n-1})(1-a_{n+1})$ is positive while $1-a_{n+1}^2$ is negative, clear contradiction. $\blacksquare$ Claim. If $a_n > 1$ and $a_{n+1} > 1$, then $a_{n+1} < \sqrt{a_n}$. Proof. Recalling that all the $a_n$ are positive, we have $0 \leq 1-a_{n+2} < 1$. Hence \[a_{n+1}^2 - 1 \leq (a_n-1)(1-a_{n+2}) < a_n - 1\]and the result follows. $\blacksquare$ Now suppose that $a_n < 1$ and $a_{n+1} > 1$ for some $n$. (Note that the first inequality must be strict.) Then \[a_{n+2} \geq \frac{a_{n+1}^2 - a_n}{1-a_n} > \sqrt{a_{n+1}}\]where the last inequality rearranges to $\frac{\sqrt{a_{n+1}} - 1}{1-a_n} > 0$. This contradicts the previous claim. This is enough to show that $\{a_n\}$ has all terms eventually at most $1$, as needed.

Does this work? I feel like I am missing something.

A crappy solution, hope it's correct. FTSOC let $a_{2022}\geq1$. Let $n=2021$ and we get $a_{2022}^{2}+a_{2021}a_{2023}\leq a_{2021}+a_{2023}$. From this we can easily get $(1-a_{2022})(1+a_{2022})\geq (1-a_{2021})(1-a_{2023})$. Since the LHS of the inequation is negative, RHS should be negative. Let $1-a_{2021}\leq1$. Then $1-a_{2023}\geq1$. Now, If $n=2020$, then just like above, we get $(1-a_{2021})(1+a_{2021})\geq (1-a_{2020})(1-a_{2022})$. So, $1-a_{2020}\geq1$. If we add the $n=2020$ and $n=2021$ case we get: $\hspace{3cm}$ $a_{2022}^2+a_{2021}^2-a_{2023}-a_{2021}-a_{2020}-a_{2022}+a_{2021}a_{2023}+a_{2020}a_{2022}\leq0 \Rightarrow$ $\hspace{3cm}$ $(a_{2022}+a_{2023})(a_{2022}-1)+(a_{2021}+a_{2020})(a_{2021}-1)\leq0$ the LHS is positive because $a_{2022}\geq1$, $a_{2021}\geq1$ and all number of the sequence is positive. Contradiction. A similar argument can be used for the other case.

Suppose FTSOC that $a_{2022}>1$. Claim: $a_n,a_{n+1}>1$ for some $n$. Proof. Suppose that $a_{2021}\le 1$. Then, $a_{2023}(1-a_{2021})\ge (a_{2022})^2-a_{2021}\implies a_{2023}>1$, as desired. $\blacksquare$ Notice if $a_n>1$, then $0<(a_n-1)a_{n+2}\le a_n-a_{n+1}^2$ so $a_n>a_{n+1}^2$. Similarly, if $a_{n+2}>1$, then $0<{a_{n+2}-1}(a_n)\le a_{n+2}-a_{n+1}^2$ so $a_{n+2}>a_{n+1}^2$. Shifting the index, we have $a_{n+1}>a_n^2$ if $a_{n+1}>0$. Since $a_n,a_{n+1}>1$ for some $n$ by our claim, we have a contradiction. $\square$

Rewrite the equation as $$(a_n-1)(a_{n+2}-1)\leq 1-(a_{n+1})^2$$Assume FTSOC that $1<a_{2022}$. Then by letting $n=2021$ we get that exactly one of $a_{2021}$ and $a_{2023}$ is larger than $1$. WLOG assume that $1<a_{2023}$. Manipulating the equation when $n=2021$ and when $n=2022$ gives $(a_{2022})^2<a_{2023}$ and $(a_{2023})^2<a_{2022}$, a contradiction.

$(a_{n+1})^2 \le a_n+a_{n+2}-a_na_{n+2}$ $(a_{n+1})^2 \le 1-(a_n-1)(a_{n+2}-1)$ Assume that $a_{2022}>1$. Then, one of $a_{2021}$ and $a_{2023}$ is greater than one, and the other is less than one. Case 1: $a_{2021}>1$, $a_{2023}<1$ Because $a_{2021}>1$, one of $a_{2020}$ and $a_{2022}$ is greater than one. So $a_{2020}<1$. As a result, $a_{2019}>1$ and $a_{2021}>1$ Then, get that $a_{2018}>1$ $a_{2017}<1$ $a_{2016}>1$ $a_{2015}>1$ $a_{2014}<1$ ... $a_1<1$ Let $a,b,c,d=a_1,a_2,a_3,a_4$ $c^2 \le 1-(b-1)(d-1)=1+(b-1)(1-d) \le 1+(b-1)(1) = b$ $b^2 \le 1-(a-1)(c-1)=1+(c-1)(1-a) \le 1+(c-1)(1) = c$ However, squaring the first equation gives $c^4 \le b^2 \le c$, which does not work, since $c > 1$. This is a contradiction. Case 2: $a_{2021}<1$, $a_{2023}>1$ Note that this is the same as the first case, but the numbers are reflected over $a_{2022}$. As a result, this is also a contradiction, so $\boxed{a_{2022} \le 1}$.

**Problem** 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 $n$. Then $a_{2022}\leq 1$. **Solution** We can rewrite the given inequality as: $$(a_{n+1})^2 \leq a_n(1 - a_{n+2})$$Since $a_n$ and $a_{n+2}$ are positive, we have: $$a_{n+1} \leq \sqrt{a_n(1 - a_{n+2})}$$ We will prove by induction that $a_n \leq 1$ for all $n \geq 1$. * **Base case:** If $a_1 > 1$, then from the inequality, we have $a_2^2 \leq a_1(1 - a_3) < 0$, which is a contradiction. Thus, $a_1 \leq 1$. * **Inductive step:** Assume $a_k \leq 1$ for some positive integer $k$. Then, $$a_{k+1}^2 \leq a_k(1 - a_{k+2}) \leq 1 \cdot (1 - 0) = 1$$Therefore, $a_{k+1} \leq 1$. By induction, we have shown that $a_n \leq 1$ for all $n \geq 1$. Since $a_{2022}$ is a member of the sequence, it follows that $a_{2022} \leq 1$.

Denote $b_{k} = a_{k} - 1$. Now rewrite the condition in terms of $b_{n}$ $$-b_{n-1}b_{n+1}\ge b_{n}(b_{n}+2)$$Now suppose that there exists $n\ge 3$ such that $b_{n}> 0$. So we get that $b_{n-1}b_{n+1} < 0$. Suppose that $b_{n-1} >0$ and $b_{n+1} < 0$ (the other case is solved similarly). From $-b_{n-2}b_{n}\ge b_{n-1}(b_{n-1}+2)$ we get that $b_{n-2} < 0$. Keep in mind that since $a_{k} > 0$ we have $b_{k} > -1$ so we get: $$b_{n-1} > -b_{n-1}b_{n+1}\ge b_{n}(b_{n}+2) \ge 2b_{n}$$$$b_{n} > -b_{n}b_{n-2}\ge b_{n-1}(b_{n-1}+2)\ge 2b_{n-1}$$ Which is a contradiction. $\square$

Suppose for every \(n\) $a_n>1$ So for any \(n\) $(a_{n+1})^2-a_n a_{n+2} \leq a_n+a_{n+2}$ $(a_{n+1})^2-a_n \leq a_{n+2} (1-a_n)$ Now as $a_n>1$ for all \(n\) $a_{n+1}<a_{n+2}$ for all \(n\) Also similarly $a_{n+1}<a_n$ Now putting \(n+1\) in place of \(n\) we can show $a_{n+2}<a_{n+1}$ Which is \(contradiction\) So there is some \(k\) $a_k \leq 1$ $(a_{k+1})^2 -a_k a_k+2 \leq a_k +a_{k+2}$ $(a_{k+1})^2 -a_k \leq a_{k+2} (1-a_k) \leq 0$ $a_{k+1} \leq 1$ For any $m>k$ $a_m\leq1$ Now putting $k-1$ in the place of \(k\) place we can prove similarly $a_{k-1} \leq 1$ so for all $m<k$ $a_m \leq 1$ so for all \(n\) $a_n \leq 1$ so $a_{2022} \leq 1$ and we are done.

The first thing we want to do is to factorize this expression. We rewrite the equation as the following. \[(1-a_n)(1-a_{n+2}) \leq 1-a_{n+1}^2\] Let $<$ denote a term that is less than or equal to $1$, and $>$ otherwise. Now assume FTSOC $a_{2022} >1$, Now looking at $a_{2021}$ and $a_{2023}$ we see that one of them must less than one, while the other will be more than one. We can see that this will give us two possibilities. 1)The terms $a_{2020}, a_{2021}, a_{2022}, a_{2023}$ being $<>><$ 2)The terms $a_{2021}, a_{2022}, a_{2023}, a_{2024}$ being $<>><$ Claim: We can't have terms $a_i,a_{i+1},a_{i+2}$ and $a_{1+3}$ with signs $<>><$. This will give us that $a_i<1$. Then we claim that doesn't work. We have. \[1-a_{i+2} < (1-a_{i})(1-a_{i+2}) \leq (1-a_{i+1})(1+a_{i+1})\]So $a_{i+1}^2<a_{i+2}$. Similarly we can get that $a_{i+2}^2<a_{1+1}$, this implies that $a_{i+1} <1$ hence a contradiction. Hence since we cant have the array $<>><$ then this implies $a_{2022}$ can't be more than $1$, hence we are done!

storage

Claim: We can get $a_i$ to be under $1$ for some $i \le 3$. Proof: If $a_{1} \le 1$ or $a_2 \le 1$, we are done. Otherwise, we can easily see that we can write $(a_2 - 1)(a_2 + 1) + (a_1 - 1)(a_3 - 1) \le 0$, so we must have $a_3 - 1$ negative as desired. Claim: Once the sequence dips below (at most) $1$, we can never get back above $1$. Proof: Take $a_n \le 1$, $a_{n + 1} > 1$, we force that the sequence cannot continue. We then get $a_{n + 2} \ge \frac{(a_{n + 1}^2 - a_n)}{1 - a_{n}}$. However, we also have $a_{n + 2}^2 + a_{n + 1}a_{n + 3} \ge a_{n + 1} + a_{n + 3}$, since $a_{n + 1}a_{n + 3} > a_{n + 3}$ we must have $a_{n + 2}^2 < a_{n + 1}$. Since $a_{n + 2}^2 \ge (\frac{(a_{n + 1}^2 - a_n)}{1 - a_{n}})^2$, we desire to show $(\frac{(a_{n + 1}^2 - a_n)}{1 - a_{n}})^2 \ge a_{n + 1}^2$, which forces a contradiction of $a_{n + 2}^2 > a_{n + 2}^2$. Indeed, rearranging we desire $1 - \frac{a_n}{a_{n + 1}^2} \ge 1 - a_{n}$ (we can remove the square because it strictly increases the RHS), but this is obviously true since $a_{n + 1} > 1$. Combining the claims gives the desired result.

Expanding, this is equivalent to, \[(a_n-1)(a_{n+2}-1)\le (1-a_{n+1})(1+a_{n+1}).\]Now, define $b_n=a_n-1$ (note $\implies b_n>-1$), substituting; \[b_nb_{n+2}=-b_{n+1}(2+b_{n+1}).\]We want to show that $b_{n+1}\le 0$ ($\iff a_{n+1}\le 1$). FTSOC, $b_{k+1}>0$. Then, $n=k$ gives \[b_kb_{k+2}=-b_{k+1}(2+b_{k+1})<0,\]meaning that either $b_k<0$ and $b_{k+2}>0$ or vice-versa. Let us first consider the former. Letting $n=k+1$, \[b_{k+1}b_{k+3}=-b_{k+2}(2+b_{k+2})>0,\]\[\implies b_{k+3}<0.\]Now, let $w=-b_n<1$, $x=b_{n+1}$, $y=b_{n+2}$, and $z=-b_{n+3}<1$. We know \[y>wy\ge x(x+2),\]\[x>xz \ge y(y+2),\]adding gives \[0>(x^2+x)+(y^2+y).\]However, \[f(x)=(x+1)(x)\ge 0\forall x>-1,\]\[\implies (x^2+x)+(y^2+y)\ge 0,\]contradiction. The other case ($b_k>0$, $b_{k+2}<0$) follows similarly. Therefore $b_{k+1}\le 0$, and $k=2021$ gives \[\implies a_{2022}\le 1,\]as desired.