I'm trying to prove the following inequality:
$$\prod_{1\leq i,j\leq n}\frac{1+a_ia_j}{1-a_ia_j}\geq1\tag{1}$$ for $a_i\in(-1,1)$. I first tried induction but doesn't seem to work well. Special cases can be proved like $n=2,3$ using brute force but I am trying to find a simpler proof. Some observations:
Define the function $f_n(x_1,x_2,\ldots,x_n)=\prod_{1\leq i,j\leq n}\frac{1+x_ix_j}{1-x_ix_j}$ for $(x_1,\ldots,x_n)\in(-1,1)^n$. Note that if $a_i\geq0$ for all $i$ then all the terms in the product are at least $1$ and hence $(1)$ is trivially true. So, is it possible to prove the following?
$$f_n(x_1,x_2,\ldots,x_n)\geq1\iff f_n(-x_1,x_2,\ldots,x_n)\geq1$$, if yes then by symmetry we can repeat this process and make all $a_i\geq0$. Also, using Induction we can do the following: Base case $n=1$ is trivial. Assuming for some $n\geq1$, we see that
$$f_n(x_1,x_2,\ldots,x_n)f_n(-x_1,x_2,\ldots,x_n)=f_{n-1}(x_2,\ldots,x_n)^2\geq1$$ and hence at least one of the following is true: $f_n(x_1,x_2,\ldots,x_n)\geq1$ or $f_n(-x_1,x_2,\ldots,x_n)\geq1$.
The case of equality
As @RiverLi pointed out and @MartinR wrote an answer, it is evident that equality holds if and only if $$\sum_{k=1}^{\infty}\frac{1}{2k-1}\left( \sum_{i=1}^n a_i^{2k-1}\right)^2=0$$The partial sums form an increasing sequence of nonnegative reals. So, the series can evaluate to zero if and only if $$\sum_{i=1}^na_i^{2k-1}=0,k\geq1$$
which is a separate and interesting problem and is discussed here.
Best Answer
Found here on AoPS:
For $-1 < x < 1$ we have the Taylor series $$ \ln (1+x) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} x^k \, , $$ which implies $$ \ln(1+x) - \ln(1-x) = 2\sum_{k=1}^\infty \frac{1}{2k-1} x^{2k-1} \, . $$ It follows that $$ \sum_{i, j=1}^n \bigl(\ln(1+a_ia_j) - \ln(1-a_ia_j)\bigr) = 2 \sum_{i, j=1}^n \sum_{k=1}^\infty \frac{1}{2k-1} a_i^{2k-1}a_j^{2k-1} \\ = 2 \sum_{k=1}^\infty \frac{1}{2k-1}\sum_{i, j=1}^n a_i^{2k-1}a_j^{2k-1} = 2 \sum_{k=1}^\infty \frac{1}{2k-1} \left( \sum_{i=1}^n a_i^{2k-1}\right)^2 \ge 0 \, . $$ This proves that the logarithm of $\prod_{1\leq i,j\leq n}\frac{1+a_ia_j}{1-a_ia_j}$ is non-negative, so that the product is $\ge 1$.
Remark: With the substitution $a_i = \tanh(x_i)$ one can see that the inequality is equivalent to $$ x_1, \ldots, x_n \in \Bbb R \implies \prod_{i, j = 1}^n \frac{\cosh(x_i + x_j)}{\cosh(x_i - x_j)} \ge 1 \, , $$ and the substitution $a_i = \tan(y_i)$ shows that it is also equivalent to $$ y_1, \ldots, y_n \in (-\frac \pi 4, \frac \pi 4) \implies \prod_{i, j = 1}^n \frac{\cos(y_i - y_j)}{\cos(y_i + y_j)} \ge 1 \, . $$