Vtyjkyuk

$d:Bbb C times Bbb C to Bbb R$ Defined by $$d(z,w) := 2fracz-wsqrt^2)(1+,$$ 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?

Lemma. Let $(E,langle,rangle)$ be an inner product space. Then, for non-zero vectors $x$ and $y$ we have
$$
leftVertfracxVert xVert^2-fracyVert yVert^2rightVert=fracVert x-yVertVert yVert Vert xVert.
$$

Proof. Indeed,
$$eqalign
left( Vert yVert Vert xVert
leftVertfracxVert xVert^2-fracyVert yVert^2rightVertright)^2&=
leftVertfracVert yVertVert xVertx-fracVert xVertVert yVertyrightVert^2cr
&=Vert yVert^2+Vert xVert^2-2Re(langle x,yrangle)cr
&=Vert x-yVert^2

$$

Corollary. Let $(H,langle,rangle)$ be an inner product space. For $(x,y)in H^2$, define
$$d(x,y)=fracVert x-yVertsqrt(1+Vert xVert^2)(1+Vert yVert^2).$$
Then $d(x,z)leq d(x,y)+d(y,z)$ for every $x,y,z$ in $H$.

Proof. Indeed, we consider $E=Htimes mathbbC$, equipped with the inner product
$$
langle (x,a),(y,b)rangle_E=langle x,yrangle_H+barab
$$
Applying the Lemma to the nonzero elements $X=(x,1)$ and $Y=(y,1)$ we see that
$$
d(x,y)=leftVert fracXVert XVert^2-fracYVert YVert^2rightVert_E
$$
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)&=leftVert fracXVert XVert^2-fracZVert ZVert^2rightVert_Ecr
&leqleftVert fracXVert XVert^2-fracYVert YVert^2rightVert_E
+leftVert fracYVert YVert^2-fracZVert ZVert^2rightVert_Ecr
&leq d(x,y)+d(y,z),

$$
and the corollary follows.

Finally, the considered question corresponds to the particular case $H=mathbbC$. because the factor $2$ is superfluous.

We need to prove that:
$$|u-v|sqrt^2le |u-w|sqrtv+|v-w|sqrtu.$$
Indeed, by C-S and the triangle inequality we obtain:
$$left(|u-w|sqrtv+|v-w|sqrturight)^2=$$
$$=|u-w|^2left(1+|v|^2right)+|v-w|^2left(1+|u|^2right)+2|(u-w)(v-w)|sqrtvgeq$$
$$geq|u-w|^2left(1+|v|^2right)+|v-w|^2(1+|u|^2)+2|(u-w)(v-w)|left(1+|uv|right)=$$
$$=left(|u-w|+|w-v|right)^2+left(|(u-w)v|+|(w-v)u|right)^2geq$$
$$geqleft(|u-w+w-v|right)^2+left(|uv-wv+wu-vu|right)^2=|u-v|^2left(1+|w|^2right).$$