Dr. Dongyi Wei is an assistant professor at the School of Mathematical Sciences of Peking University. He recently proposed the below exercise that GPT-4 fails to solve. I propose in this post two solutions, a direct one and another one based on functional analysis.

Exercise. Let $\{a_1, \dots, a_n\} \subseteq (-1, 1)$ be given. Show that $$\prod_{1 \le i, j \le n} \frac{1 + a_i a_j}{1 - a_i a_j} \ge 1.$$

Direct proof#

Remark that we only need to prove that $$\sum_{1 \le i, j \le n} \log ( 1 + a_i a_j ) - \log ( 1 - a_i a_j ) \ge 0.$$ Recall that for all $t \in (-1, 1)$, we have $$\log(1 + t) = \sum_{k = 1}^{\infty} \frac{(-1)^{k + 1} t^k}{k}.$$ Therefore, we have $$\begin{aligned} \sum_{1 \le i, j \le n} \log ( 1 + a_i a_j ) - \log ( 1 - a_i a_j ) & = \sum_{1 \le i, j \le n} \sum_{k = 1}^{\infty} \frac{(-1)^{k + 1} a_i^k a_j^k}{k} + \frac{a_i^k a_j^k}{k}\\ & = \sum_{1 \le i, j \le n} \sum_{k = 0}^{\infty} \frac{2 a_i^{2 k + 1} a_j^{2 k + 1}}{2 k + 1}\\ & = \sum_{k = 0}^{\infty} \frac{2}{2 k + 1} \sum_{1 \le i, j \le n} a_i^{2 k + 1} a_j^{2 k + 1}\\ & = \sum_{k = 0}^{\infty} \frac{2}{2 k + 1} \bigg( \sum_{i = 1}^n a_i^{2 k + 1} \bigg)^2 \ge 0. \end{aligned}$$ The last equality is the key to the proof, although it is not obvious at first sight.

Proof based on functional analysis#

This proof has been proposed to me by Dr. Zaikun Zhang. It reformulates the above problem into a (more general) $2$-dimensional problem. We first prove the following theorem.

Theorem. Let $f$ be an integrable function on $\mathbb{R}$ with $\lVert f \rVert_{\infty} < 1$. Then $$\iint_{\mathbb{R}^2} \log \big( 1 + f ( x ) f ( y ) \big) \mathrm{d} x \mathrm{d} y \ge \iint_{\mathbb{R}^2} \log \big( 1 - f ( x ) f ( y ) \big) \mathrm{d} x \mathrm{d} y$$

Proof. In a similar way as in the direct proof, since $\lVert f \rVert_{\infty} < 1$, for $( x, y ) \in \mathbb{R}^2$, we have $$\log \big( 1 + f ( x ) f ( y ) \big) - \log \big( 1 - f ( x ) f ( y ) \big) = \sum_{k = 0}^{\infty} \frac{2 f^{2k + 1} ( x ) f^{2k + 1} ( y )}{2k + 1}.$$ Therefore, the integrability of $f$ and the fact that $\lVert f \rVert_{\infty} < 1$ provide $$\begin{aligned} \iint_{\mathbb{R}^2} \big[ \log \big( 1 + f ( x ) f ( y ) \big) - \log \big( 1 - f ( x ) f ( y ) \big) \big] \mathrm{d} x \mathrm{d} y & = \sum_{k = 0}^{\infty} \frac{2}{2k + 1} \iint_{\mathbb{R}^2} f^{2k + 1} ( x ) f^{2k + 1} ( y ) \mathrm{d} x \mathrm{d} y\\ & = \sum_{k = 0}^{\infty} \frac{2}{2k + 1} \bigg( \int_{\mathbb{R}} f^{2k + 1} ( x ) \mathrm{d} x \bigg)^2 \ge 0, \end{aligned}$$ which concludes the proof.

The solution to the original problem is a straightforward consequence of the above theorem, with $$f(x) = \begin{cases} a_1 & \text{if $0 \le x < 1$,}\\ \vdots\\ a_n & \text{if $n - 1 \le x < n$,}\\ 0 & \text{otherwise.} \end{cases}$$