$V$ is inner product space. $u, v \in V$ are two orthogonal vectors.
Prove that $\|v-u\| \geq \|v\|$.
Because $\|v-u\|, \|v\| \geq 0$ it's enough to prove that $||v-u||^2 \geq \|v\|^2$.
$$\|v-u\|^2 = \langle v, v \rangle - \underbrace{\langle u,v \rangle}_0 - \underbrace{\langle v,u \rangle}_0 + \langle u, u \rangle = \|v\|^2 + \|u\|^2.$$
Which means that I need to show that $||u||^2 \geq 0$, and that's given by definition.
I'm not sure if this correct.
Thanks in advance.
You basically have it, but $\|u\|^2 \ge 0$ is not "given by definition". Rather, it follows from the fact that the square of a real number is nonnegative, and $\|u\|$ is a real number . It would be just fine if you said that $\|v\|^2 + \|u\|^2 \ge \|v\|^2$ since $\|u\|^2 \ge 0$.