2025 International Mathematical Olympiad China National Team Selection Test

Proof 1: Let $\Lambda$ be the set consisting of $\pm {\lambda }_i$ for $1\le i\le 2025$. For positive integer $n$, define functions $S,K:{\Lambda }^{2n}\to {\mathbb{Z}}_{\ge 0}$ as: $$S(c_1,\dots ,c_{2n})=\#\{(x_1,\dots ,x_{2n})\in X^{2n}\mid c_1{x}_1+\cdots +c_{2n}{x}_{2n}=0\},$$ $$K(c_1,\dots ,c_{2n})=\#\{1\le i\le 2n\mid c_i\in \{\pm c_1\}\}.$$ Let $T_{2n}$ be the cardinality of: $$\{(x_1,\dots ,x_{2n})\in X^{2n}\mid x_1+\cdots +x_n=x_{n+1}+\cdots +x_{2n}\},$$ then $T_{2n}=S(c_1,\dots ,c_1,-c_1,\dots ,-c_1)$.

Lemma: The maximum value of function $S$ is $T_{2n}$.

Proof of Lemma: Assume the maximum value of $S$ is $M\ge 1$, and let $(c_1,\dots ,c_{2n})$ be a point in $S^{-1}(\{M\})$ where $K$ attains its maximum. Let $K(c_1,\dots ,c_{2n})=k$, and assume $c_i\in \{\pm c_1\}$ for $1\le i\le k$. For real $y$, define: $$I_1(y)=\#\{(x_1,\dots ,x_n)\in X^n\mid c_1{x}_1+\cdots +c_n{x}_n=y\},$$ $$I_2(y)=\#\{(x_{n+1},\dots ,x_{2n})\in X^n\mid -c_{n+1}{x}_{n+1}-\cdots -c_{2n}{x}_{2n}=y\}.$$ Then: $$S(c_1,\dots ,c_{2n})=\sum _y{I}_1(y)I_2(y).$$ By Cauchy-Schwarz: $$\begin{array}{rl}{M}^2& =S(c_1,\dots ,c_{2n}{)}^2\le \sum _y{I}_1(y{)}^2\sum _y{I}_2(y{)}^2\\ & =S(c_1,\dots ,c_n,-c_1,\dots ,-c_n)\cdot S(c_{n+1},\dots ,c_{2n},-c_{n+1},\dots ,-c_{2n})\\ & \le M^2,\end{array}$$ implying $S(c_1,\dots ,c_n,-c_1,\dots ,-c_n)=M$. By maximality of $K$: $$k=K(c_1,\dots ,c_{2n})\ge \min \{2k,2n\},$$ thus $k=2n$. Therefore all $c_i\in \{\pm c_1\}$, and: $$M=T_{2n}.$$ This completes the lemma's proof.

Returning to the main problem, let $n=1013$. For real $y$, define: $$J_1(y)=\#\{(x_1,\dots ,x_n)\in X^n\mid c_1{x}_1+\cdots +c_n{x}_n=y\},$$ $$J_2(y)=\#\{(x_{n+1},\dots ,x_{2n-1})\in X^{n-1}\mid -c_{n+1}{x}_{n+1}-\cdots -c_{2n-1}{x}_{2n-1}=y\}.$$ Similarly using Cauchy-Schwarz: $$\begin{gathered} A^2={\left(\sum _y{J}_1(y)J_2(y)\right)}^2\le \sum _y{J}_1(y{)}^2\sum _y{J}_2(y{)}^2 \\ \le S(c_1,\dots ,c_n,-c_1,\dots ,-c_n)\cdot S(c_{n+1},\dots ,c_{2n-1},-c_{n+1},\dots ,-c_{2n-1}). \end{gathered}$$ Combining with the lemma yields: A^2 \le T_{2026} \cdot T_{2024} = B \cdot C. \quad \square **Proof 2:** (Based on solutions by Deng Leyan and Zhang Hengye) For non-zero real $p$, using Newton-Leibniz formula: \lim_{T \to +\infty} \frac{1}{T} \int_0^T e^{ipt} dt = 0. $$Thus:$$ \lim_{T \to +\infty} \frac{1}{T} \int_{0}^{T} e^{ipt} dt = $$\{\begin{array}{ll}0,& p\ne 0,\\ 1,& p=0.\end{array}$$ Let $f(t) = \sum_{x \in X} e^{ixt}$. Then: |A| = \lim_{T \to +\infty} \frac{1}{T} \left| \int_{0}^{T} e^{-ibt} \prod_{j=1}^{2025} f(\lambda_j t) dt \right|. $$ByH\ddot{o}lde{r}^{\prime }sinequality:$$ $$\begin{array}{rl}|A|& =\underset{T\to +\infty }{\lim }\frac{1}{T}\left|{\int }_0^T{e}^{-ibt}\prod _{j=1}^{2025}f({\lambda }_{j}t)dt\right|\\ & \le \underset{T\to +\infty }{\lim }\frac{1}{T}{\int }_0^T\prod _{j=1}^{2025}|f({\lambda }_{j}t)|dt\\ & =\underset{T\to +\infty }{\lim }\frac{1}{T}{\int }_0^T\prod _{j=1}^{2025}(|f({\lambda }_{j}t){|}^{2025}{)}^{\frac{1}{2025}}dt\\ & \le \underset{T\to +\infty }{\lim }\prod _{j=1}^{2025}{\left(\frac{1}{T}{\int }_0^T|f({\lambda }_{j}t){|}^{2025}dt\right)}^{\frac{1}{2025}}\\ & =\prod _{j=1}^{2025}{\left(\underset{T\to +\infty }{\lim }\frac{1}{T}{\int }_0^T|f(|{\lambda }_j|t){|}^{2025}dt\right)}^{\frac{1}{2025}}\\ & =\prod _{j=1}^{2025}{\left(\underset{T\to +\infty }{\lim }\frac{1}{|{\lambda }_j|T}{\int }_0^{|{\lambda }_j|T}|f(s){|}^{2025}ds\right)}^{\frac{1}{2025}}\\ & =\underset{T\to +\infty }{\lim }\frac{1}{T}{\int }_0^T|f(t){|}^{2025}dt.\end{array}$$ $$Similarly:$$ |B| = \lim_{T \to +\infty} \frac{1}{T} \int_{0}^{T} |f(t)|^{2024} dt, |C| = \lim_{T \to +\infty} \frac{1}{T} \int_{0}^{T} |f(t)|^{2026} dt. $$Finally,byCauchy-Schwarz:$$ $$\begin{array}{rl}|B|\cdot |C|& =\underset{T\to +\infty }{\lim }\frac{1}{T^2}{\int }_0^T|f(t){|}^{2024}dt{\int }_0^T|f(t){|}^{2026}dt\\ & \ge \underset{T\to +\infty }{\lim }\frac{1}{T^2}{\left({\int }_0^T|f(t){|}^{2025}dt\right)}^2\\ & ={\left(\underset{T\to +\infty }{\lim }\frac{1}{T}{\int }_0^T|f(t){|}^{2025}dt\right)}^2\\ & \ge |A{|}^2,\end{array}$$ $$