Showing that two vectors are not orthogonal, using Cauchy-Schwarz

Question

The problem is to show that the vectors $u$ and $u - v$ are not orthogonal, knowing that $\|u\| > \|v\|$.

The definition says that $u$ and $u - v$ are orthogonal if and only if $\langle u, u - v \rangle = 0$. I am trying to show that $\langle u, u - v \rangle$ is bigger than $0$ using Cauchy-Schwarz inequality, that says that $|\langle u, v \rangle| \le \|u\| \|v\|$ for all $u$ and $v$ in a vector space $V$ (that what the textbook gives as a hint). I have no idea where to start, and it would be very appreciating if someone could solve this problem and explain their solution.

Answer 1

Use $\langle u,u-v\rangle=\langle u,u\rangle-\langle u,v\rangle \ge\|u\|^2-\|u\|\|v\|$ because of Cauchy-Schwarz

Answer 2

First, your CS inequality is missing a square.

We have $$|\langle u,u-v\rangle|^2 \geq ||u||\cdot || u-v|| \geq ||u||(||u||-||v||)>0$$

using CS and the triangle inequality for the norm.

Answer 3

We have

$$\left|\langle u, u-v\rangle\right| = \left|\|u\|^2 - \langle u, v\rangle \right| \ge \left|\|u\|^2 - \left|\langle u, v\rangle\right|\right| \ge \|u\|^2 - \left|\langle u, v\rangle\right| \ge \|u\|^2 - \|u\|\|v\| > 0$$

because $\|u\| > \|v\|$.