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

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.

• Mar 27, 2023 at 10:44
• See: (Chinese) mp.weixin.qq.com/s/_FC4nDxOq9aZgwa2VIqYcg Mar 27, 2023 at 10:44
• @RiverLi: Btw, with the substitution $a_i = \tanh(x_i)$ an equivalent formulation of this inequality is $\prod_{i, j = 1}^n \frac{\cosh(x_i + x_j)}{\cosh(x_i - x_j)} \ge 1$ for arbitrary real numbers $x_1, \ldots, x_n$. I wonder if that gives more insight into the problem or not. Apr 6, 2023 at 7:02
• Also with $a_i = \tan(y_i)$ it becomes $\prod_{i, j = 1}^n \frac{\cos(y_i - y_j)}{\cos(y_i + y_j)} \ge 1$ for $y_1, \ldots, y_n \in (-\pi/4, \pi/4)$. Apr 6, 2023 at 7:05
• @MartinR Interesting. I don't have another approach now. Apr 6, 2023 at 7:19

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

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 .

• How do you determine $x_1, \ldots, x_n$ such that $x_i+x_j=xa_ia_j$ holds for all $i$ and $j$? I do not think that is always possible. Apr 3, 2023 at 14:33
• @MartinR $\pm e^{\ln(a)+\ln(b)}=\pm ab$ Apr 3, 2023 at 14:39
• I don't yet believe it. Let us take a concrete example: $x=1$ and $(a_1, a_2, a_3) = (1, 2, 3)$. Can you give me three values $(x_1, x_2, x_3)$ such that $x_i + x_j = a_i a_j$ holds for $i=1, 2, 3$ and for $j = 1, 2, 3$? Apr 3, 2023 at 14:42
• You have to show that there are numbers $b_j$ and $x_j$ such that $b_ib_j\exp(x_i+x_j)=\exp(xa_ia_j)-\exp(-xa_ia_j)$ holds for all $i, j$. Perhaps it is obvious to you, but not to me. Apr 3, 2023 at 17:42

Sketch of proof :

Using Frullani's integral we have :

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

For $$b\in(-1,1),y>1$$

We can show the sum of exponential for fixed $$x$$ on $$x\in(u,v)$$ and then using $$y$$ to have the whole positive $$x$$-axis .

The sum is :

$$S(x)=\sum_{i,j=1}^{n}\sinh\left(a_{i}a_{j}x\right)$$

First lemma

Moreover we have the inequality for $$z,y\in(-1,1)$$ and $$\exists x\in(u,v),u>0,0 and $$0<\varepsilon<1$$ then :

$$e^{xyz}-e^{-xyz}-2xyz+\varepsilon x^{2}y^{2}z^{2}\geq 0\tag{k}$$

We can show $$k$$ in substituting $$U=xyz$$ we have :

$$f(U)=2\sinh\left(U\right)-2U+\varepsilon U^{2}$$

Then :

$$f'(U)=2\varepsilon U+2\cosh(U)-2$$

Setting $$x=-a=-\varepsilon$$ we have :

$$f'(a)=\cosh(a)-1+a^2\geq 0$$

Setting $$x=-a$$ we have :

$$f(-a)\geq 0$$

So we have on $$x\in (0,a)$$ :

$$J=\int_{0}^{a}\frac{e^{-x}S(x)-e^{-x}\left(I\right)}{x}\geq0$$

$$I=2x\left(\sum_{i=1}^{n}a_{i}\right)^{2}-x^2\varepsilon\left(\sum_{i=1}^{n}a_{i}^{2}\right)^{2}$$

But using the property of Frullani's integral $$J$$ an as for $$t>1$$ $$(a_1a_j,\cdots,a_na_n,a_1a_jt,\cdots,a_na_nt)$$ majorize $$(-a_1a_j,\cdots,-a_na_n,-a_1a_jt,\cdots,-a_na_nt)$$ is also $$t=\varepsilon$$:

$$0\leq J\leq-\int_{0}^{\varepsilon }\frac{e^{-x}I}{x}dx+ \int_{0}^{t}\frac{e^{-\frac{x}{t}}S\left(\frac{x}{t}\right)}{x}dx=\int_{0}^{1}\frac{e^{-x}S\left(x\right)}{x}dx+-\int_{0}^{\varepsilon }\frac{e^{-x}I}{x}dx$$

For the other part we can substitute $$x=1/X$$ and use :

$$\int_{t}^{\infty}\frac{e^{-\frac{1}{t}x}S\left(\frac{x}{t}\right)}{x}dx=\int_{1}^{\infty}\frac{e^{-x}S\left(x\right)}{x}dx$$

Now making $$\varepsilon\to 0$$ we have the result .