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

Let $(a_i)$ be a sequence of real numbers that satisfy $0 < |a_i| < 1,i=1,\ldots,n$. Then we want to prove that $$\sum_{i,j=1}^n \frac{a_ia_j}{1 - a_i^2a_j^2}\geq 0.$$

I know how to solve this using the geometric series: since $a_i^2a_j^2 < 1$ we have using the geometric series that $$ \sum_{i,j=1}^n \frac{a_ia_j}{1 - a_i^2a_j^2} = \sum_{i,j=1}^n a_ia_j\sum_{k=0}^\infty(a_ia_j)^{2k}=\sum_{k=0}^\infty\sum_{i,j=1}^na_i^{2k+1}a_j^{2k+1}=\sum_{k=0}^\infty\left(\sum_{i=1}^na_i^{2k+1}\right)^2\geq 0. $$

My question is whether we can solve this using another method that does not involve the geometric series.

My first approach is to somehow split the denominator into two terms involving $a_i$ and $a_j$ individually, so that I can turn the sum into a quadratic form. I can see that $1-a_i^2a_j^2 = (1 - a_ia_j)(1 + a_ia_j)$, but this sort of splitting is not helpful since each term contains both $a_i$ and $a_j$, and is antisymmetric. So, I'm wondering if there even exist $b_i,b_j>0$ such that $b_ib_j = 1 - a_i^2a_j^2$ for all $i,j$.

Another approach would perhaps be to represent the fraction as an integral? Or maybe even represent the denominator as a double integral. Involvement of complex numbers can also be an option.

I personally think that there isn't a better, cleaner way to solve this, but it's been bothering me for quite some time now.

V.S.e.H.
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    this version seems a bit funny. A braindead strong induction yields a slightly weaker inequality:

    $$n\sum_{i=1}^{n+1}\dfrac{a_i^2}{1-a_i^4} + (n-1)\sum_{i\neq j}^{n+1}\dfrac{a_ia_j}{1-a_i^2a_j^2}\geq 0,$$ which makes me think a smarted induction solution probably exists.

     – dezdichado Apr 07 '23 at 23:17
  • Also, it seems that the function $f(x) = \dfrac{x}{1-x^2}$ satisfies the following: $$x_1,x_2,\ldots x_m\in (-1,1),, x_1 +x_2 +\ldots+x_m \geq 0\implies \sum_{i=1}^m f(x_i)\geq 0.$$ Certainly true for $n=2,3$ and just stuck on one part of the induction currently. If proven, this is much more general because it will only require $f$ to be odd, monotonically increasing and convex on $[0,1)$ (and necessarily concave on $(-1,0]$) – dezdichado Apr 09 '23 at 15:13
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    @dezdichado, If $(x_1,x_2,x_3)=(-\frac{2}{3},\frac{1}{3},\frac{1}{3})$, then $$x_1+x_2+x_3=0\qquad\text{but}\qquad f(x_1)+f(x_2)+f(x_3)=-\frac{9}{20}.$$ – Sangchul Lee Apr 09 '23 at 15:34
  • hmmm never mind then. – dezdichado Apr 09 '23 at 16:07
  • @dezdichado It seems that any function $f$ that satisfies your conditions has $\sum_{i,j=1}^n f(a_ia_j) \geq 0$, i.e. any inverse sigmoid. The question is how to exploit these properties without resorting to Taylor series. – V.S.e.H. Apr 09 '23 at 20:54
  • Oh yeah. I have essentially showed that it comes to checking a certain Karamata condition about majorization. I just can't figure out a succinct way of writing it. – dezdichado Apr 09 '23 at 23:18
  • @dezdichado Post an answer maybe? – V.S.e.H. Apr 10 '23 at 00:22
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    V.S.e.H. I am planning to - but even if I manage to write it within a reasonable length, I just don't think it will be more elegant than @Sangchul Lee's brilliant solution. On a personal note, I think the best solution is using the geometric series argument. – dezdichado Apr 10 '23 at 01:00

1 Answers1

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Let $|b_i| < 1$, $i=1,\ldots,n$ and define the quadratic form $Q$ on $\mathbb{R}^n$ by

$$ Q(\mathbf{x}) = \sum_{i,j=1}^{n} \frac{x_i x_j}{1 - b_i b_j}. $$

Also, let $T = \operatorname{diag}(b_1, \ldots, b_n)$. Then

$$ Q(\mathbf{x}) = \left( \sum_{i=1}^{n} x_i \right)^2 + Q(T\mathbf{x}) \geq Q(T\mathbf{x}). \tag{*} $$

From this, we get

$$ Q(\mathbf{x}) \geq Q(T^k\mathbf{x}) \xrightarrow{k\to\infty} Q(\mathbf{0}) = 0, $$

showing that $Q$ is positive semi-definite. Now OP's case follows by plugging $b_i = a_i^2$ and $x_i = a_i$ into $Q(\mathbf{x})$.

Remark. This is essentially the proof using geometric series in disguise. So I am not fully satisfied with this.

Sangchul Lee
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