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

Let $$u, v \in \mathbb R^n$$. Prove that if $$\|u+tv\| \ge \|u\|$$ for all $$t \in \mathbb R$$, then $$u\cdot v=0$$ (vectors $$u$$ 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.

$$\|u\|^{2}+t^{2}\|v\|^{2}+2t \langle u, v \rangle \geq \|u\|^{2}$$ for all $$t$$. This gives $$t^{2}\|v\|^{2}+2t \langle u, v \rangle \geq 0$$. Take $$t >0$$, divide by $$t$$ and let $$t \to 0$$ You get $$\langle u, v \rangle \geq 0$$. If you take limit through negative values you get the reverse inequality.

$$||u+tv||^2\geq ||u||^2$$, so $$f(t)=||u+tv||^2$$ has global -- and therefore also local -- minimum at $$t=0$$, hence has $$f'(0)=2 u\cdot v=0$$. (This is actually Kavi's answer "in disguise".)

Although the original question was interested only in the case where the vector space of interest is $$\Bbb R^n$$, the existing answer has shown the result is more general than that, applying to any vector space over $$\Bbb R$$. In fact, we can generalise even further, a point I think is of some interest: if in a space over $$\Bbb C$$ the inequality runs over all complex $$t$$ (or $$z$$, as I'll call it to make the complexity manifest), the orthogonality still follows.

Suppose $$V$$ is a vector space over $$\mathbb{C}$$ and $$u,\,v\in V$$ such that $$\forall z\in\mathbb{C}\left(\left\Vert u+zv\right\Vert \ge\left\Vert u\right\Vert \right)$$; then $$\left\langle u|v\right\rangle =0$$. We'll assume the inner product is antilinear in its right argument. Define $$x:=\Re z,\,y:=\Im z$$ and $$a:=\Re\left\langle u|v\right\rangle ,\,b:=\Im\left\langle u|v\right\rangle$$ so $$0\le\left\Vert u+zv\right\Vert ^{2}-\left\Vert u\right\Vert ^{2}=\left(x^{2}+y^{2}\right)\left\Vert v\right\Vert ^{2}-2\left(ax-by\right)$$for all $$x,\,y\in\mathbb{R}$$. The case $$x=\frac{a}{\left\Vert v\right\Vert ^{2}},\,y=\frac{-b}{\left\Vert v\right\Vert ^{2}}$$ gives $$\left(x^{2}+y^{2}\right)\left\Vert v\right\Vert ^{2}-2\left(ax-by\right)=-\frac{a^{2}+b^{2}}{\left\Vert v\right\Vert ^{2}},$$so $$a=b=0$$. Hence $$\left\langle u|v\right\rangle =0$$.