a_507_bc wrote: Let $a_1, a_2, \ldots, a_{2023}$ be positive reals such that $\sum_{i=1}^{2023}a_i^i=2023$. Show that $$\sum_{i=1}^{2023}a_i^{2024-i}>1+\frac{1}{2023}.$$ Subtitute $x_i=a_i^i$. We want to minimize $S:=\sum_{i=1}^{2023}x_i^{\frac{2024-i}{i}}$ for non-negative reals $x_1,\ldots,x_{2023}$ (not positive reals as in the problem statement!) with $\sum_{i=1}^{2023}x_i=2023$. We have \[ \frac{\partial S}{\partial x_k}=\frac{\partial}{\partial x_k}\left(x_k^{\frac{2024-k}{k}}\right)=\frac{2024-k}{k}x_k^{\frac{2024-2k}{k}} \]In particular $\frac{\partial S}{\partial x_k}$ is strictly increasing for $1\leq k\leq 1011$, constantly $1$ for $k=1012$ and strictly increasing for $1013\leq k\leq2023$. For $1\leq k\leq 1011$ we have $\frac{\partial S}{\partial x_k}\geq1$ for $x_k\geq\left(\frac{k}{2024-k}\right)^{\frac{k}{2024-2k}}$. So if there is an index $k\leq1011$ with $x_k>\left(\frac{k}{2024-k}\right)^{\frac{k}{2024-2k}}$ we can replace $(x_k,x_{1012})$ with $\left(\left(\frac{k}{2024-k}\right)^{\frac{k}{2024-2k}},x_{1012}+x_k-\left(\frac{k}{2024-k}\right)^{\frac{k}{2024-2k}}\right)$ making $S$ smaller in the process. So we can assume that $x_k\leq\left(\frac{k}{2024-k}\right)^{\frac{k}{2024-2k}}<1$ for $k\leq1011$. This implies $\sum_{i=1012}^{2023}x_i>1012$. If there are different indices $i,j\geq1012$ with $x_i,x_j>0$ assume w.l.o.g. $\frac{2024-i}{i}x_i^{\frac{2024-2i}{i}}\leq \frac{2024-j}{j}x_j^{\frac{2024-2j}{j}}$. We have \[ (x_i+x_j)^{\frac{2024-i}{i}}=x_i^{\frac{2024-i}{i}}+\int_{x_i}^{x_i+x_j}\frac{2024-i}{i}x^{\frac{2024-2i}{i}}\;\mathrm{d}x<x_i^{\frac{2024-i}{i}}+\int_0^{x_j}\frac{2024-j}{j}x^{\frac{2024-2j}{j}}\;\mathrm{d}x=x_i^{\frac{2024-i}{i}}+x_j^{\frac{2024-j}{j}}. \]So we can assume there is exacly one index $i\geq1012$ with $x_i>0$. Since $C:=\sum_{i=1012}^{2023}x_i>1012$ we have $x_i=C>1012$. The term $C^{\frac{2024-i}{i}}$ is minimal for $i=2023$ when choosing an index $i\in\{1012,\ldots,2023\}$ such that $x_i=C$ and all other indices are $0$. So we can assume $x_{2023}>1012$. But now we have \[ S>x_{2023}^{\frac{1}{2023}}>1012^{\frac{1}{2023}}>e^{\frac{1}{2023}}>1+\frac{1}{2023}. \]

This problem was proposed by me. Of course there is a much shorter solution, and without any analysis.

can you post your solution ? i think solution about taking minimum element of set and do some inequality but i can't finish

Well, the official solutions can already be found online, but I can also describe it here: Let us prove conversely that if $a_1^{2023}+a_2^{2022}+\dots+a_{2023} \le 1+\frac{1}{2023}$, then $a_1+a_2^2+\dots+a_{2023}^{2023}<2023$. This is trivial if all $a_i$ are all less than $1$, as then every summand is also less than $1$. Moreover, by assumption at most one variable can be $\ge 1$ since otherwise the first sum is already at least $2$. So we are left with the case where exactly one variable is $\ge 1$, say $a_i \ge 1$. Now just notice that for the second half of the indices i.e. $j \ge 1012$ with $j \ne i$ we have $a_j^{j} \le a_j^{2024-j}$, hence the total contribution of these terms is $\le \frac{1}{2023}$ by assumption. Also, the terms $j \le 1011$ with $j \ne i$ contribute at most $1$ each, hence at most $1011$ in total. Finally, the term $i$ itself contributes at most $\left(1+\frac{1}{2023}\right)^{2023}$. Hence the total sum is at most $1011+\frac{1}{2023}+\left(1+\frac{1}{2023}\right)^{2023}$ which is far more than enough (we can use the classical fact that the last summand is at most $e<3$, but even a much weaker estimate would suffice).

Here we present an award for the least homogenized inequality ever created.