Prove that this is a metric

Question

$d:\Bbb C \times \Bbb C \to \Bbb R$ Defined by $$d(z,w) := 2\frac{|z-w|}{\sqrt{(1+|z|^2)(1+|w|^2) }},$$ prove that $d$ is metric in $\Bbb C$.

I had proved $d$ satisfies the two conditions to be metric.. I do not know how to prove the triangular inequality. Can anyone help me?

Answer 1

Lemma. Let $(E,\langle,\rangle)$ be an inner product space. Then, for non-zero vectors $x$ and $y$ we have $$ \left\Vert\frac{x}{\Vert x\Vert^2}-\frac{y}{\Vert y\Vert^2}\right\Vert=\frac{\Vert x-y\Vert}{\Vert y\Vert \Vert x\Vert}. $$

Proof. Indeed, $$\eqalign{ \left( \Vert y\Vert \Vert x\Vert \left\Vert\frac{x}{\Vert x\Vert^2}-\frac{y}{\Vert y\Vert^2}\right\Vert\right)^2&= \left\Vert\frac{\Vert y\Vert}{\Vert x\Vert}x-\frac{\Vert x\Vert}{\Vert y\Vert}y\right\Vert^2\cr &=\Vert y\Vert^2+\Vert x\Vert^2-2\Re(\langle x,y\rangle)\cr &=\Vert x-y\Vert^2 } $$

Corollary. Let $(H,\langle,\rangle)$ be an inner product space. For $(x,y)\in H^2$, define $$d(x,y)=\frac{\Vert x-y\Vert}{\sqrt{(1+\Vert x\Vert^2)(1+\Vert y\Vert^2)}}.$$ Then $d(x,z)\leq d(x,y)+d(y,z)$ for every $x,y,z$ in $H$.

Proof. Indeed, we consider $E=H\times \mathbb{C}$, equipped with the inner product $$ \langle (x,a),(y,b)\rangle_E=\langle x,y\rangle_H+\bar{a}b $$ Applying the Lemma to the nonzero elements $X=(x,1)$ and $Y=(y,1)$ we see that $$ d(x,y)=\left\Vert \frac{X}{\Vert X\Vert^2}-\frac{Y}{\Vert Y\Vert^2}\right\Vert_E $$ Thus, for $x,y,z$ from $H$, then, (with notation $X=(x,1)$, $Y=(y,1)$ and $Z=(z,1)$,) we have $$\eqalign{ d(x,z)&=\left\Vert \frac{X}{\Vert X\Vert^2}-\frac{Z}{\Vert Z\Vert^2}\right\Vert_E\cr &\leq\left\Vert \frac{X}{\Vert X\Vert^2}-\frac{Y}{\Vert Y\Vert^2}\right\Vert_E +\left\Vert \frac{Y}{\Vert Y\Vert^2}-\frac{Z}{\Vert Z\Vert^2}\right\Vert_E\cr &\leq d(x,y)+d(y,z), } $$ and the corollary follows.

Finally, the considered question corresponds to the particular case $H=\mathbb{C}$. because the factor $2$ is superfluous.