combination of points in the open unit disk also lie in the unit disk

Question

Suppose $a,b$ are points in the open unit disk $\{z\in\mathbb{C}:|z|<1\}$. Then, does the combination $$\frac{(1-|a|^2)b+(1-|b|^2)a}{1-|ab|^2}$$ also lie in the unit disk (open)?

I think yes, but am unable to prove. The triangle inequality does not give any hopes here. I think Cauchy-Schwarz should be used in some way here. Any hints how to proceed? Thanks beforehand.

Answer 1

Using the triangle inequality, the absolute value of the expression is at most $$ \frac{(1-|a|^2)|b|+(1-|b|^2)|a|}{1-|ab|^2} = \frac{(|a|+|b|)(1-|a||b|)}{1-|a|^2 |b|^2} = \frac{|a|+|b|}{1+|a||b|} \\ = 1 - \frac{(1-|a|)(1-|b|)}{1+|a||b|} < 1 \, . $$

Remark: This is related to Conformal automorphism of unit disk that interchanges two given points. If $a, b$ are distinct points in the unit disk then $$ T(z) = \frac{c- z}{1- \bar cz} $$ with $$ c = \frac{(1-|a|^2)b+(1-|b|^2)a}{1-|ab|^2} $$ is the (unique) automorphism of the unit disk with $T(a) = b$ and $T(b) = a$.

Answer 2

Let $t=|a|, s=|b|$. If we show that $(1-t^{2})s+(1-s^{2})t<1-t^{2}s^{2}$ we are done.

This can be written as $s-st^{2}+t-s^{2}t+t^{2}s^{2} <1$. To prove this inequality let us show that the left side is an increasing function of $s$ in $[0,1]$ whose value at $s=1$ is $1-t^{2}+t-t+t^{2}=1$.

The derivative w.r.t. $s$ is $1-t^{2}-2ts+2st^{2}$. Prove that this is positive by observing that $1+t >2t >2ts$. [ Multiply this last inequality by $1-t$].