Prove that this is a metric $d:\Bbb C \times \Bbb C \to \Bbb R$ 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?
 A: 
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.
A: We need to prove that:
$$|u-v|\sqrt{1+|w|^2}\le |u-w|\sqrt{1+|v|^2}+|v-w|\sqrt{1+|u|^2}.$$
Indeed, by C-S and the triangle inequality we obtain:
$$\left(|u-w|\sqrt{1+|v|^2}+|v-w|\sqrt{1+|u|^2}\right)^2=$$
$$=|u-w|^2\left(1+|v|^2\right)+|v-w|^2\left(1+|u|^2\right)+2|(u-w)(v-w)|\sqrt{\left(1+|v|^2\right)\left(1+|u|^2\right)}\geq$$
$$\geq|u-w|^2\left(1+|v|^2\right)+|v-w|^2(1+|u|^2)+2|(u-w)(v-w)|\left(1+|uv|\right)=$$
$$=\left(|u-w|+|w-v|\right)^2+\left(|(u-w)v|+|(w-v)u|\right)^2\geq$$
$$\geq\left(|u-w+w-v|\right)^2+\left(|uv-wv+wu-vu|\right)^2=|u-v|^2\left(1+|w|^2\right).$$
