Prove that $\prod_{1\leq i,j\leq n}\frac{1+a_ia_j}{1-a_ia_j}\geq1$ for $n$ real numbers $a_i\in(-1,1)$

Question

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.

Answer 1

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 $\ge 1$.

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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 $$ y_1, \ldots, y_n \in (-\frac \pi 4, \frac \pi 4) \implies \prod_{i, j = 1}^n \frac{\cos(y_i - y_j)}{\cos(y_i + y_j)} \ge 1 \, . $$

Answer 2

Extended comment :

We have two definitions of exponentialy convex function :

A function is called exponentially convex if :

$$g\left(t\right)=e^{f\left(ta+\left(1-t\right)b\right)}-\left(1-t\right)e^{f\left(b\right)}-te^{f\left(a\right)}\leq 0,\forall a,b\in I, t\in[0,1]$$

We have the equivalent definition if $f(x)$ is continuous:

$$\sum_{i,j=1}^{n}b_{i}b_{j}f\left(x_{i}+x_{j}\right)\geq 0$$

$\forall n\geq 1,\forall b_i\in R,x_i\in I$

Now it's not hard to show that $f(x)=\exp(cx)$ ,$c\in(-1,1),a,b\in(-\infty,\infty)$ is exponential convex because in the first definition it's less than zero .

So plugging ,$b_ib_j\exp(x_i+x_j)=\exp(xa_ia_j)-\exp(-xa_ia_j)$ and using the second definition we have :

$$\sum_{i,j=1}^{n}\sinh{a_ia_jx}\geq 0\tag{I}$$

Now using Frullani's integrals we have :

$$\int_{0}^{\infty}\frac{e^{-\left(1-a_{i}a_{j}\right)x}-e^{-\left(1+a_{i}a_{j}\right)x}}{x}dx=\ln\frac{1+a_{i}a_{j}}{1-a_{i}a_{j}}$$

Now summing we have $(\operatorname{I})$ with a factor .

We have shown it's positive or zero so the inequality is established .