Baltic Way 2023 Shortlist

Let us prove that conversely, the condition $$a_1^{2023}+a_2^{2022}+\cdots +a_{2023}\le 1+\frac{1}{2023}$$ implies that $$S:=a_1+a_2^2+\cdots +a_{2023}^{2023}<2023.$$ This is trivial if all $a_i$ are less than $1$. So suppose that there is an $i$ with $a_i\ge 1$, clearly it is unique and $a_i<1+\frac{1}{2023}$. Then we have $$\begin{array}{rl}{a}_i^i& <{\left(1+\frac{1}{2023}\right)}^{2023}=1+\sum _{k=1}^{2023}\frac{1}{k!}\cdot \frac{2023}{2023}\cdot \frac{2022}{2023}\cdots \cdot \frac{2023-k+1}{2023}\\ & <1+\sum _{k=1}^{2023}\frac{1}{k!}\le 1+\sum _{k=0}^{2022}\frac{1}{2^k}<3,\end{array}$$ $$\sum _{\begin{array}{c}k=1,\\ k\ne i\end{array}}^{1011}{a}_k^k\le 1011\text{and}\sum _{\begin{array}{c}k=1012,\\ k\ne i\end{array}}^{2023}{a}_k^k\le \sum _{\begin{array}{c}k=1012,\\ k\ne i\end{array}}^{2023}{a}_k^{2024-k}<\frac{1}{2023}.$$ Hence we have $$S=a_i^i+\sum _{\begin{array}{c}k=1,\\ k\ne i\end{array}}^{1011}{a}_k^k+\sum _{\begin{array}{c}k=1012,\\ k\ne i\end{array}}^{2023}{a}_k^k<3+1011+\frac{1}{2023}<2023.$$