An alternative lower bound for $\prod_{ i,j = 1}^n\frac{1+a_ia_j}{1-a_ia_j} $

Question

In Prove that $\prod_{1\leq i,j\leq n}\frac{1+a_ia_j}{1-a_ia_j}\geq1$ for $n$ real numbers $a_i\in(-1,1)$ it was shown that for real numbers $a_1, \ldots, a_n \in (-1, 1)$ we always have $$ P = \prod_{ i,j = 1}^n\frac{1+a_ia_j}{1-a_ia_j} \ge 1 \, . $$ The proof (originally from AoPS) uses the Taylor series of the logarithm to derive an explicit formula for the logarithm of that product as an infinite sum of squares with positive coefficients: $$ \tag{*} \ln P = 2 \sum_{k=1}^\infty \frac{1}{2k-1} \left( \sum_{i=1}^n a_i^{2k-1}\right)^2 \ge 0 \, . $$

If we omit all terms for $k \ge 2$ on the right-hand side then we get the weaker inequality $$ \tag 1 \ln P \ge 2 ( a_1 + a_2 + \cdots + a_n)^2 $$ or equivalently $$ \tag 2 P \ge e^{2( a_1 + a_2 + \cdots + a_n)^2 } $$

My question: Is there a simpler/more direct way to obtain $(1)$ or $(2)$ without the use of infinite series?

We cannot use $$ \ln(1+a_ia_j) - \ln(1-a_ia_j) \ge 2 a_i a_j $$ because that holds only if $a_ia_j \ge 0$. Another idea is to consider the function $$ f(x) = \prod_{ i,j = 1}^n\frac{1+x^2a_ia_j}{1-x^2a_ia_j} \, . $$ for $0 \le x \le 1$. From the representation $(*)$ we know that $\ln f(x)$ is increasing in $x$, but it is not obvious (to me) how to prove that directly, since the terms in $$ \frac{d}{dx} \ln f(x) = \sum_{i, j=1}^n \frac{4a_i a_j x}{1-a_i^2a_j^2 x^4} $$ can be both positive and negative.

Answer 1

We have the base case of $1+a_1^2 \ge 1-a_1^2$ for $a_1 \in (-1,1)$

Now suppose we have $P = \prod_{ (i,j) \in A} (1+a_ia_j) \ge \prod_{ (i,j) \in A} (1-a_ia_j) \, $ for some nonempty $A \subsetneq \{1..n\}\times\{1..n\}$

Let $(i_0,j_0),(i_1,j_1) \in \{1..n\}\times\{1..n\} - A$,

Now

$$\begin{aligned}P &= \prod_{ (i,j) \in A} (1+a_ia_j) \\ &= \sum_{k=0}^{|A|} \sum_{S \subset A}^{|S| = k} C_S \prod_{ (i,j) \in S} a_ia_j\\ \end{aligned}$$

Where $C_S = \{ \# $ ways to get a permute of $S$ from $ A\}$

And

$$\begin{aligned}Q &= \prod_{ (i,j) \in A} (1-a_ia_j) \\ &= \sum_{k=0}^{|A|} \sum_{S \subset A}^{|S| = k} (-1)^k\ C_S \prod_{ (i,j) \in S} a_ia_j\\ \end{aligned}$$

Then $$\begin{aligned}\frac{P-Q}{2} &= \sum_{\substack{ k=1 \\ k\ odd }}^{|A|} \sum_{S \subset A}^{|S| = k} C_S \prod_{ (i,j) \in S} a_ia_j \end{aligned}$$

Calling the above expression $I(A)$, our inequality is equivalent to $I(A) \ge 0$

Now considering the inductive case $A' = A\cup \{(i_0,j_0)\}\cup \{(i_1,j_1)\}$

$$\begin{align}I(A') &= \sum_{\substack{ k=1 \\ k\ odd }}^{|A|+2} \sum_{\substack{ S \subset A &\cup \{(i_0,j_0)\}\\ &\cup \{(i_1,j_1)\} }}^{|S| = k} C_S \prod_{ (i,j) \in S} a_ia_j \\\\ &= \sum_{\substack{ k=1 \\ k\ odd }}^{|A|} \ \sum_{\substack{T \subset \{&(i_0,j_0),&(i_1,j_1)\} }} \ \Biggl( \biggl[\prod_{ (i,j) \in T} a_ia_j \biggr] \Biggl( \ \sum_{S \subset A }^{|S| = k} \biggl[ C_S \prod_{ (i,j) \in S} a_ia_j \biggr] \ \Biggr) \Biggr) \end{align}$$

In other words

$$\begin{align} I(A') &= I(A) + I(A \cup \{(i_0,j_0)\}) + I(A \cup \{(i_1,j_1)\}) + I(A \cup \{(i_0,j_0), (i_1,j_1)\})\\ &= I(A) + a_{i_0}a_{j_0}I(A) + a_{i_1}a_{j_1}I(A) + a_{i_0}a_{j_0}a_{i_1}a_{j_1}I(A) \\ &= (1 + a_{i_0}a_{j_0} + a_{i_1}a_{j_1} + a_{i_0}a_{j_0}a_{i_1}a_{j_1})I(A) \\ &= (1 + a_{i_0}a_{j_0})(1 + a_{i_1}a_{j_1})I(A) \\ &\ge 0 \end{align}$$

And we are done.

Answer 2

Draft :

Let $-1<x<1$ then define :

$$f\left(x\right)=\ln\left(\frac{1+x}{1-x}\right)$$

then we have :

$$f'''\left(x\right)>0$$

Then the idea is to use Levinson's inequalities see https://rgmia.org/papers/v15/v15a68.pdf

Example case $n=2$ we have $x,a\in(-1,1),c=0 $:

$$f\left(a^{2}+c\right)+f\left(b^{2}+c\right)+2f\left(ab+c\right)-(f\left(-a^{2}-c\right)+f\left(-b^{2}-c\right)+2f\left(-ab-c\right))\geq 4f\left(\frac{\left(a^{2}+b^{2}+2ab+3c\right)}{4}\right)-4f\left(\frac{-\left(a^{2}+b^{2}+2ab+3c\right)}{4}\right)\geq 2(a+b)^2$$

Some explanation :

As we have equality (3) in the linked paper combined to $\ln\left(\frac{1+x}{1-x}\right)=-\ln\left(\frac{1-x}{1+x}\right)$

We can switch the problematic values (not the squares) and the proof of theorem 4 is the same .

Remains to show the last hand side .

We are done .

Edit :

Fuch 's inequality works because it the vector (a^2,b^2,2ab) majorize the vector in the other side.But the argument above is still incomplete