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.
On
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 $\left\langle u|v\right\rangle =0$.
$\|u\|^{2}+t^{2}\|v\|^{2}+2t \langle u, v \rangle \geq \|u\|^{2}$ 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.