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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.

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    $\begingroup$ See: artofproblemsolving.com/community/… $\endgroup$
    – River Li
    Mar 27, 2023 at 10:44
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    $\begingroup$ See: (Chinese) mp.weixin.qq.com/s/_FC4nDxOq9aZgwa2VIqYcg $\endgroup$
    – River Li
    Mar 27, 2023 at 10:44
  • $\begingroup$ @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. $\endgroup$
    – Martin R
    Apr 6, 2023 at 7:02
  • $\begingroup$ 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)$. $\endgroup$
    – Martin R
    Apr 6, 2023 at 7:05
  • $\begingroup$ @MartinR Interesting. I don't have another approach now. $\endgroup$
    – River Li
    Apr 6, 2023 at 7:19

3 Answers 3

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

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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 .

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  • $\begingroup$ 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. $\endgroup$
    – Martin R
    Apr 3, 2023 at 14:33
  • $\begingroup$ @MartinR $\pm e^{\ln(a)+\ln(b)}=\pm ab$ $\endgroup$
    – DesmosTutu
    Apr 3, 2023 at 14:39
  • $\begingroup$ 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 $j = 1, 2, 3$? $\endgroup$
    – Martin R
    Apr 3, 2023 at 14:42
  • $\begingroup$ 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. $\endgroup$
    – Martin R
    Apr 3, 2023 at 17:42
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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 $\exists x\in(u,v),u>0,0<u<v<1$ 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 .

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