# Cauchy-Schwarz inequality and angle between two vectors

Notes

I am reading these notes, and I can't understand the Cauchy-Schwarz inequality.

It says that it proves that the input is between $$[-1,1]$$. The Cauchy-Schwarz inequality only states that the product between two vectors divided over the product of their distances is less than $$1$$ not greater than or equal to $$-1$$.

For example:

$$\frac{\langle x,y\rangle}{\|x\|\|y\|}\leq 1.$$

But $$\frac{\langle x,y\rangle}{\|x\|\|y\|}$$ could return $$-10$$ for some values of $$x$$ and $$y$$?

• The CS inequality states that the absolute value of that expression is at most one. See for example here May 24, 2022 at 7:35
• $$\frac{|\langle x,y\rangle|}{\|x\|\|y\|}\leq 1.$$ May 24, 2022 at 7:36

## 1 Answer

The Cauchy-Schwarz inequality states that $$|\langle x,y\rangle| \le \Vert x\Vert \cdot \Vert y \Vert$$ holds for all vectors $$x, y$$ in an inner product space. Therefore, if both $$x$$ and $$y$$ are not zero, $$-1 \le \frac{\langle x,y\rangle}{\Vert x\Vert \cdot \Vert y \Vert} \le 1$$ and the angle between these vectors can be defined via $$\theta = \arccos \frac{\langle x,y\rangle}{\Vert x\Vert \cdot \Vert y \Vert} \, .$$

Remark: If you “know” the Cauchy-Schwarz inequality only as $$\langle x,y\rangle \le \Vert x\Vert \cdot \Vert y \Vert$$ without the absolute value on the left-hand side then $$-\langle x,y\rangle = \langle - x,y\rangle \le \Vert -x\Vert \cdot \Vert y \Vert = \Vert x\Vert \cdot \Vert y \Vert \\ \implies \langle x,y\rangle \ge - \Vert x\Vert \cdot \Vert y \Vert$$ gives you the estimate in the other direction.