# Prove that $\sum_{i,j=1}^n \frac{a_ia_j}{1 - a_i^2a_j^2}\geq 0$ where each $|a_i| < 1$

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.

• 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. Commented Apr 7, 2023 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 $(-1,0]$) Commented Apr 9, 2023 at 15:13
• @dezdichado, If $(x_1,x_2,x_3)=(-\frac{2}{3},\frac{1}{3},\frac{1}{3})$, then $$x_1+x_2+x_3=0\qquad\text{but}\qquad f(x_1)+f(x_2)+f(x_3)=-\frac{9}{20}.$$ Commented Apr 9, 2023 at 15:34
• hmmm never mind then. Commented Apr 9, 2023 at 16:07
• 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. Commented Apr 10, 2023 at 1:00

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.

• (+1) Still it was rather ingenious. Commented Apr 7, 2023 at 18:25
• This proof looks slick, +1! I'll wait for awhile for other answers to pop up and accept if not. Commented Apr 7, 2023 at 18:32