Vectors: If $||u+tv|| \ge ||u||$ for all $t$, prove that $u\cdot v=0$

Let $$u$$ and $$v$$ be vectors in $$\mathbb R^n$$. The exercise is to prove that if $$||u+tv|| \ge ||u||$$ for all real $$t$$, then $$u\cdot v=0$$ ($$u$$ and $$v$$ are perpendicular).

I tried writing $$v$$ as $$(n+xu)$$, where $$u\cdot n=0$$, and then try to prove that $$x$$ must be zero, but was unable to develop this solution.

Proof: $U$ is an ultrafilter of a Boolean algebra $B$ if and only if for all $x$ in $U$ exactly one of $x$,$x^*$ belongs to $U$.

I have been stuck with this problem for a while now. I have a proof that letting $$U$$ be an ultrafilter, exactly one of $$x,x^*$$ belongs to $$U$$ for all $$x$$ in $$B$$, I did this by showing that both belong to $$U$$ implies that $$U=B$$ which cannot be the case, and if neither long to $$U$$ then a contradiction can be derived by de-Morgan’s laws. However, I’m stuck with the reverse implication, I need to prove that:

“If $$U$$ is a subset of a Boolean algebra $$B$$ such that $$\forall x \in B$$, exactly one of $$x \in U$$ and $$x^* \in U$$ is true.

Is this even the case? Any help would be greatly appreciated, thanks in advance.