Claim: Suppose $u,v$ ∈ V. Then $<u,v>$ $=$ $0$ iff $||u|| $ ≤ $||u+av||$ ∀ $a$ ∈ $F$
Proof. ( ⟹) Suppose $<u,v>$ $=$ $0$. By properties of the inner product we have that ∀ $a$ ∈ $F$ $<u,av>$ $=$ conjugate of a $<u,v>$. So $<u,av>$ $=$ $0$ since $<u,v>$ $=$ $0$. Now since $u$ ⊥ $av$ ⟹ $||u||^2$ $+$ $||av||^2$ $=$ $||u+av||^2$ by the Pythagorean Theorem. Thus $||u||^2$ ≤ $||u+av||^2$ ⟹ $||u||$ ≤ $||u+av||$ as claimed.
I'm stuck on the second part of the proof where I have to prove the other direction.