Show that if $z\mathbin\bot F$ then $\|z\|\leq \Vert z-y\Vert$ for every $y\in F$

Question

Let $H$ be a pre-Hilbert space and $F$ a subspace of $H.$ Then if $z \in H$ the following are equivalent:

$i) \ \forall y \in F \ \Vert z\Vert\leq \Vert z-y\Vert$

$ii) \ z$ is orthogonal to $F.$

I had no problems showing that $ii) \Rightarrow i).$ But how do I show the other implication? It looks very easy but I can't seem to able to prove it. Any hints?

Answer 1

From i) we get

$$\Vert z\Vert ^2\le \Vert z-y\Vert^2=\Vert z\Vert^2+\Vert y\Vert^2-2\langle z,y\rangle$$so $$2\langle z,y\rangle\le \Vert y\Vert ^2,\quad \forall y\in F$$ so for all $t\in\Bbb R$: $$2t\langle z,y\rangle\le t^2\Vert y\Vert ^2,\quad \forall y\in F$$ hence the reduced discriminant $\Delta'=\langle z,y\rangle^2$ for the quadratic equation is non positive and then equal to $0$. Conclude.