How to use Triangle inequality to find the projection onto unit ball?

Question

The projection onto the unit ball $$C:=\mathbb{B}(0,1)=\{x:||x||\leq1\}$$ is given by $$P_{C}(x)=\frac{x}{max\{||x||,1\}}, \quad\forall x\in X$$ where $X$ is Hilbert space. Now I can understand this but I want to use triangle inequality to show that $\frac{x}{||x||}$ is the closest point to the $\mathbb{B}(0,1)$ when $||x||\geq1$ (mabe multivalued) unless the norm is strictly convex. I am not able to built this argument using triangle inequality so if anyone can help me with this or can do this using a triangle inequality I will be very thankful to him.

Answer 1

It seems the following.

Suppose that there exist points $x$ and $y$ in a normed space $X$ such that $\|x\|\ge 1$, $y\in C$ and $$\|x-y\|<\|x-P_C(x)\|.$$ Then Triangle inequality implies $$\|x\|\le \|x-y\|+\|y\|<\|x-P_C(x)\|+1=\|x-P_C(x)\|+\|P_C(x)\| =\|x\|,$$ a contradiction.