Why do we need the polarization identity to show the simple relation $U:H\to H$ is unitary if and only if $||Ux|| = ||x||$ for all $x \in H$

Question

An operator $U:H\to H$ is unitary if and only if $\|Ux\| = \|x\|$ for all $x \in H$. I read that the proof that $\|Ux\| = \|x\|$ implies $U$ is unitary depends on the polarization identity. But it seems more elementary than that; can we not just say that $$ \|Ux\| = \|x\| \Longleftrightarrow (Ux,Ux)^\frac{1}{2} = (x,x)^\frac{1}{2} \Longleftrightarrow (Ux,Ux) = (x,x), $$ and therefore $U$ is a unitary operator. Why do we need the polarization identity?

Answer 1

I hope this theorem helps:

Theorem: If $T \in L(H)$ where $H$ is a Hilbert space then the following are equivalent:

$1)T^*T=I$

$2)<Tx,Ty>=<x,y>$ forall $x,y \in H$

$2)||Tx||=||x||$ forall $x \in H$

$Proof$

$1) \Rightarrow 2)$

If $T^*T=I$ then $<T^*Tx,y>=<x,y>, \forall x,y \in H \Rightarrow <Tx,Ty>=<x,y>, \forall x,y \in H$

$2) \Rightarrow 3)$

For $y=x$ we have that $||Tx||^2=||x||^2 \Rightarrow ||Tx||=||x||$

$3) \Rightarrow 1)$

Forall $x \in H$

$<Tx,Tx>=<x,x> \Rightarrow <T^*Tx,x>=<x,x>=<Ix,x> \Rightarrow <(T^*T-I)x,x>=0 \Rightarrow T^*T-I=0 \Rightarrow T^*T=I$