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 $y$?

  • 2
    $\begingroup$ The CS inequality states that the absolute value of that expression is at most one. See for example here $\endgroup$
    – Martin R
    May 24, 2022 at 7:35
  • 1
    $\begingroup$ $$ \frac{|\langle x,y\rangle|}{\|x\|\|y\|}\leq 1.$$ $\endgroup$ May 24, 2022 at 7:36

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .