International Mathematical Olympiad China National Team Selection Test, 2025

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:
$$S(c_1, \dots, c_{2n}) = \#\{(x_1, \dots, x_{2n}) \in X^{2n} \mid c_1 x_1 + \dots + c_{2n} x_{2n} = 0\},$$
$$K(c_1, \dots, c_{2n}) = \#\{1 \le i \le 2n \mid c_i \in \{\pm c_1\}\}.$$
Let $T_{2n}$ be the cardinality of:
$$\{(x_1, \dots, x_{2n}) \in X^{2n} \mid x_1 + \dots + x_n = x_{n+1} + \dots + x_{2n}\},$$
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:
$$I_1(y) = \#\{(x_1, \dots, x_n) \in X^n \mid c_1 x_1 + \dots + c_n x_n = y\},$$
$$I_2(y) = \#\{(x_{n+1}, \dots, x_{2n}) \in X^n \mid -c_{n+1}x_{n+1} - \dots - c_{2n}x_{2n} = y\}.$$
Then:
$$S(c_1, \dots, c_{2n}) = \sum_y I_1(y) I_2(y).$$
By Cauchy-Schwarz:
$$\begin{align*} M^2 &= S(c_1, \dots, c_{2n})^2 \le \sum_y I_1(y)^2 \sum_y I_2(y)^2 \\ &= S(c_1, \dots, c_n, -c_1, \dots, -c_n) \cdot S(c_{n+1}, \dots, c_{2n}, -c_{n+1}, \dots, -c_{2n}) \\ &\le M^2, \end{align*}$$
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:
$$J_1(y) = \#\{(x_1, \dots, x_n) \in X^n \mid c_1 x_1 + \dots + c_n x_n = y\},$$
$$J_2(y) = \#\{(x_{n+1}, \dots, x_{2n-1}) \in X^{n-1} \mid -c_{n+1}x_{n+1} - \dots - c_{2n-1}x_{2n-1} = y\}.$$
Similarly using Cauchy-Schwarz:
$$A^2 = \left(\sum_y J_1(y)J_2(y)\right)^2 \le \sum_y J_1(y)^2 \sum_y J_2(y)^2 \\ \le S(c_1, \dots, c_n, -c_1, \dots, -c_n) \cdot S(c_{n+1}, \dots, c_{2n-1}, -c_{n+1}, \dots, -c_{2n-1}).$$
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 = $$\begin{cases} 0, & p \ne 0, \\ 1, & p = 0. \end{cases}$$
$$Let \$f(t) = \sum_{x \in X} e^{ixt}\$. Then:$$
|A| = $\lim_{T \to +\infty} \frac{1}{T} \left| \int_{0}^{T} e^{-ibt} \prod_{j=1}^{2025} f(\lambda_j \right.$ t) dt $\left. \right|.$
$$By Hölder's inequality:$$
$$\begin{align*} |A| &= \lim_{T \to +\infty} \frac{1}{T} \left| \int_0^T e^{-ibt} \prod_{j=1}^{2025} f(\lambda_j t) dt \right| \\ &\le \lim_{T \to +\infty} \frac{1}{T} \int_0^T \prod_{j=1}^{2025} |f(\lambda_j t)| dt \\ &= \lim_{T \to +\infty} \frac{1}{T} \int_0^T \prod_{j=1}^{2025} (|f(\lambda_j t)|^{2025})^{\frac{1}{2025}} dt \\ &\le \lim_{T \to +\infty} \prod_{j=1}^{2025} \left( \frac{1}{T} \int_0^T |f(\lambda_j t)|^{2025} dt \right)^{\frac{1}{2025}} \\ &= \prod_{j=1}^{2025} \left( \lim_{T \to +\infty} \frac{1}{T} \int_0^T |f(|\lambda_j|t)|^{2025} dt \right)^{\frac{1}{2025}} \\ &= \prod_{j=1}^{2025} \left( \lim_{T \to +\infty} \frac{1}{|\lambda_j|T} \int_0^{|\lambda_j|T} |f(s)|^{2025} ds \right)^{\frac{1}{2025}} \\ &= \lim_{T \to +\infty} \frac{1}{T} \int_0^T |f(t)|^{2025} dt. \end{align*}$$
$$Similarly:$$
|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, by Cauchy-Schwarz:$$
$$\begin{align*} |B| \cdot |C| &= \lim_{T \to +\infty} \frac{1}{T^2} \int_0^T |f(t)|^{2024} dt \int_0^T |f(t)|^{2026} dt \\ &\ge \lim_{T \to +\infty} \frac{1}{T^2} \left( \int_0^T |f(t)|^{2025} dt \right)^2 \\ &= \left( \lim_{T \to +\infty} \frac{1}{T} \int_0^T |f(t)|^{2025} dt \right)^2 \\ &\ge |A|^2, \end{align*}$$
$$$$