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I ask for a proof of :

Let $x,y\in(-0.9,0.9)$ then it seems we have :

I have conjecture for and $x,y\in(-1,1)$ :

$$f(x,y)=\ln\left(\frac{(1+xy)^{2}(1+x^{2})}{(1-xy)^{2}(1-x^{2})}\right)-\left(x\operatorname{arctanh}(x)-y\operatorname{arctanh}(y)+6\left(xy\right)-6\left(xy\right)^{2}\right)\geq 0$$

It's an attempt to solve the problem here Prove that $\prod_{1\leq i,j\leq n}\frac{1+a_ia_j}{1-a_ia_j}\geq1$ for $n$ real numbers $a_i\in(-1,1)$ .

The only thing I know is the function for $|y|<1/2$ fixed is convex on the domain above .

Then the equality case is a limit at $x=y=0$ which is equal to zero .

If true it remains to summing the inequalities and then it's not hard to show the remainders is positive .

My strategy :

First step :

Show for $x\in(-0.9,0.9)$ and $0.8>a>0$ a fixed unknow then :

$$h(x)=f(x,a),h''(x)>0$$

Second steps :

As the function $h(x)$ is convex on that domain above we use tangent line method we have :

$$h'\left(a\left(1-a\right)\right)\left(x-a\left(1-a\right)\right)+h\left(a\left(1-a\right)\right)\leq f(x,a)$$

So it remains to show :

$$h'\left(a\left(1-a\right)\right)>0,h\left(a\left(1-a\right)\right)>0$$

A similar strategy works for $-0.8<a<0$

We have also the stronger statement $x,y\in(-0.9,0.9)$:

$$g\left(x,y\right)=\ln\left(\frac{(1+xy)^{2}(1+x^{2})}{(1-xy)^{2}(1-x^{2})}\right)-\left(x\operatorname{arctanh}(x)-y\operatorname{arctanh}(y)+6xy\right)\geq 0$$

Complement to the strategy :

We have :

Let $0<x<1$ then we have :

$$\operatorname{arctanh}\left(x\right)>2\arcsin\left(x\right)-x$$

Using the inequality above we introduce :

$$m\left(x,y\right)=\ln\left(\frac{(1+xy)^{2}(1+x^{2})}{(1-xy)^{2}(1-x^{2})}\right)-\left(x\left(2\arcsin\left(x\right)-x\right)-y\left(2\arcsin\left(y\right)-y\right)+6xy\right)$$

Then we differentiate and using $\arcsin\left(x\right)>x$ on the same domain squaring both side then the derivative is :

$$j\left(x,y\right)=-\left((2x)/\sqrt{1-x^{2}}\right)^{2}+\left((4(x^{4}(-y)-x^{3}y^{2}+x+y))/((x^{4}-1)(x^{2}y^{2}-1))-6y\right)^{2}$$

Then we use a form of Buffalo's way we have :

$$j\left(\frac{1}{2}+\frac{\frac{1}{2}x}{x+y},\frac{1}{2}+\frac{\frac{1}{2}y}{x+y}\right)\geq 0$$

Where in this example all the coefficients are positives.

How to (dis)prove it ?

Any progress is very welcome !

Barackouda
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    @Philipp Nice observation can you make an answer ? – Barackouda Mar 29 '23 at 09:56
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    A remark: Equality holds in $\prod_{ i,j = 1}^n\frac{1+a_ia_j}{1-a_ia_j} \ge 1$ if the non-zero $a_i$ come in pairs of same absolute value and opposite sign, e.g. for $(a_1, a_2, a_3) = (-0.8, 0, 0.8)$ or $(a_1, a_2, a_3, a_4) = (-0.9, -0.3, 0.3, 0.9)$. Which means that estimating the log of the product from below by sums of $f(a_i, a_j)$ terms might not work if $f(x, y)$ is strictly positive for $x \ne y$. – Martin R Mar 29 '23 at 12:19
  • @MartinR I think there is a hidden majorization here no? – Barackouda Mar 29 '23 at 14:49

1 Answers1

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Have you tried computing a minimum of this function and examine convexity on the domain? I have not verified by computing it precisely, but when plotting the function, it seems that a minimum is reached at $(x,y)=(0,0)$ and that the hessian is positive definite. In that case $0$ would indeed be a minimum. But I have not verified it by hand. More precisely, if you compute $Df|_{(0,0)}$ and $Hf|_{(0,0)}$ and if $Df|_{(0,0)} =0$ and $Hf|_{(0,0)}$ is positive definite on the domain (under the assumption that bothe exist and $f$ is $C^2$), then $0$ should be a local minimum on the domain.

I added a plot below.

Plot of the function1

Note: The computation of $Df$ and of $Hf$ would take a lot aof time. If you want to compute it exactly to verify the result, you may want to use a calculator such as Wolfram.

PNM
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