This is the coolest inequality I’ve seen in a while.

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(Dongyi Wei) Let $a_1,...,a_n\in (-1,1)$. Show that $$\prod_{1\le i,j\le n} \frac{1+a_ia_j}{1-a_ia_j}\ge 1$$ and determine all equality cases.

Heuristics

Can we smooth this? It’s unlikely - we can’t really adjust the $a_i$’s in any way that keeps most of the expression the same, because all the terms are functions of $(a_ia_j)$, and we can’t change one without changing the rest.

The other thing to do is start to think about equality cases. We get one for free: $(0,0,...,0)$. In general, notice that by setting $a_n=0$ we get the same inequality with one less variable. This suggests that with a nontrivial equality case somewhere we start getting very weird sets of equality cases.

Now we try some small cases:

• $n=1$ is just $(1+x^2) / (1-x^2) \ge 1$.
• Now we try $n=2$: the inequality is $$\p{\frac{1+x^2}{1-x^2}} \p{\frac{1+y^2}{1-y^2}} \p{\frac{1+xy}{1-xy}}^2 \ge 1.$$ By expanding, we get $$(x+y)^2(1-x^2y^2) \ge 0.$$ As we’ve feared, we have a weird equality case, where $y=-x$. We see how this works: with $y=-x$, $xy=-x^2$, so the $xy$ terms are the reciprocal of the $x^2$ and $y^2$ terms. So, the equality cases consist of (at least) $(x,-x,0,...,0)$ up to permutation.

Upon a little more staring, it seems that in general, if we had a pair $(x,-x)$ among the $n$-variables, we can just remove them, since for some other variable $z$, $$\frac{1+xz}{1-xz} \cdot \frac{1+(-x)z}{1-(-x)z} = 1$$ so the $(x,z)$ and $(-x,z)$ terms exactly cancel!

The magic step

At this point, it’s pretty clear that smoothing-type methods won’t work any more. There has to be some crazy way to directly tackle the problem to produce the equality case.

Here’s a thought: $\frac{1+x}{1-x}$ reminds me of the identity $$\tanh^{-1} x = \log \frac{1+x}{1-x}.$$ This is of great importance in statistical physics: suppose we had two states $\{\pm 1\}$, and the probability of being $1$ was proportional to $e^{\eta}$ and the probability of being $-1$ was proportional to $e^{-\eta}$. Then, the mean $m$ and the exponent $\eta$ is related by $$m = \frac{e^{\eta} - e^{-\eta}}{e^{\eta} + e^{-\eta}} = \tanh \eta$$ and to reverse it we have $$\eta = \tanh^{-1} m = \log\frac{1+m}{1-m}.$$

Another reason why this is nice: taking log turns multiplication into addition. Then, instead of a gigantic product expansion, it becomes separated series expansions in each $a_ia_j$, which is potentially more manageable.

Indeed, using the expansion $$\tanh^{-1} (x) = x + \frac{x^3}{3} + \frac{x^5}{5} + \cdots$$ we simply have $$\sum_{1\le i,j \le n} \tanh^{-1}(a_ia_j) = \sum_{k=1,3,5,...} \frac{1}{k} (a_1^k + ... + a_n^k)^2$$ so it’s clearly non-negative! Furthermore, equality only holds when all odd power sums are 0: that’s why stuff pairs up. There are two ways to see this:

• invoke the theory of linear recurrences and deduce that if $$a_1 (a_1^2)^{k} + ... + a_n(a_n^2)^{k} = 0$$ for all $k$, then we conclude that an all-zero linear coefficients must have coefficients zero, so the squaring must have collapsed together two coefficients of equal magnitude but opposite signs.
• directly argue as follows: sort the terms by magnitude in descending order, then $a_1$ has the largest maganitude (wlog $a_1$): $$1 + \p{\frac{a_2}{a_1}}^k + ... = 0$$ then taking $k\to \infty$ we see that the only possibility is that there were an equal number of $a_1$ and $-a_1$’s. Then, we remove them and repeat the argument.