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.

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    $\begingroup$ One remark: the proof only uses the fact that $\log(1+x)-\log(1-x)$ has nonnegative Maclaurin series coefficients (and that the series has radius of convergence at least $1$, I guess), and so the result holds with $\log(1+x)-\log(1-x)$ replaced by any function $f(x)$ with the same property. $\endgroup$ Mar 29 at 13:06
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    $\begingroup$ @GregMartin More generally, a series with radius of convergence $R$ allows for any $a_i$ of modulus less than $\sqrt{R}$. $\endgroup$
    – J.G.
    Mar 29 at 14:22
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    $\begingroup$ An "obvious" reason why it is true: - if exactly $k$ of the $n$ numbers are negative then there are $k^2+(n-k)^2$ factors that are $\geq 1$ and $2k(n-k)$ are $\leq 1$ and since AM-GM says $k^2+(n-k)^2\geq 2k(n-k),$ the whole product is $\geq 1. $ $\endgroup$
    – dezdichado
    Apr 7 at 23:33
  • $\begingroup$ @dezdichado Following your remark can we show it using math.stackexchange.com/questions/4244874/… $\endgroup$ Apr 10 at 14:04
  • $\begingroup$ @ErikSatie I don't see how that will be applicable. Essentially, for an odd, increasing on $f$ that is convex on$[0,1)$ and concave on $(-1,0],$ we can use Karamata to conclude if we can show that it is possible to extract $2k(n-k)$ of the positive terms that can majorize the negatives. I checked manually for $n=3,4$ but it's ugly, case-by-case solution. $\endgroup$
    – dezdichado
    Apr 10 at 16:28

2 Answers 2


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


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

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


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

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.

  • $\begingroup$ I think induction should work. But why is your particular case of $A'$ to have two more elements than $A$ ? Also, I think OP is looking for a stronger lower bound than just $1$ $\endgroup$
    – dezdichado
    Apr 7 at 13:30
  • $\begingroup$ $I(A')$ is a sum over odd subsets, adding two elements preserves parity, kind of, I am sweeping under the rug the fact that the middle terms $a_{i_0}a_{j_0}I(A)$ and $a_{i_1}a_{j_1}I(A)$ in the recurrence are actually sums over even subsets. A fully rigorous proof would need induction on both functions. The lower bound of 1 can't be done away with entirely since it is met when the $a_i$ are $0$, but this proof, when corrected, constructs the precise value of their difference. $\endgroup$
    – Zane
    Apr 7 at 19:39
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    $\begingroup$ Is this a proof of $P = \prod_{ i,j = 1}^n\frac{1+a_ia_j}{1-a_ia_j} \ge 1 $ or a proof of $ P \ge e^{2( a_1 + a_2 + \cdots + a_n)^2 }$? – The latter is what my question is about: “Is there a simpler/more direct way to obtain (1) or (2) ...” $\endgroup$
    – Martin R
    Apr 10 at 13:35

Draft :

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


then we have :


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

  • $\begingroup$ We can tackle it with $a_ia_j-c_{i,j}-a_ia_j+c_{i,j}=0,c_{i,j}>0$ in rearranging the positive and the negative $a_ia_j$ such that $\min(\cdots)\leq \max(\cdots)$ and we can use theorem 2 $\endgroup$ Apr 10 at 9:22
  • $\begingroup$ @MartinR I show the case $n=2$ can you proceed further ? $\endgroup$ Apr 10 at 14:34
  • $\begingroup$ @MartinR I wait you .It works due to the symmetry we can remove the constraint $\max\leq \min $ . $\endgroup$ Apr 10 at 17:19

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