---
tag: 0RFWW56Y22
kind: statement
role: theorem
name: Cauchy–Schwarz inequality
msc: [15A63, 46C05]
uses: [01EK923SKP, 0KG4D8AP6G]
sources:
  - url: https://en.wikipedia.org/wiki/Cauchy–Schwarz_inequality
    locator: Statement of the inequality
same_as:
  wikidata: Q190546
  wikipedia: Cauchy–Schwarz inequality
---

## Statement

Let $V$ be an [[inner product space|0QVNCJXRY3]] over $\R$ or $\C$, with the [[induced norm|0MFJJV97RJ]] $\norm{\cdot}$.
Then, for all $x, y \in V$,
$$\abs{\langle x, y \rangle} \le \norm{x} \norm{y}.$$
Equality holds if and only if $x$ and $y$ are linearly dependent.

## Remarks

For the dot product on $\R^n$, it says $\left(\sum_i x_i y_i\right)^2 \le \left(\sum_i x_i^2\right)\left(\sum_i y_i^2\right)$.
In a real inner product space, it allows the definition of the angle $\theta \in [0, \pi]$ with $\cos\theta = \langle x, y \rangle / (\norm{x} \norm{y})$ for nonzero $x$ and $y$.
One proof uses that $\norm{x - ty}^2 \ge 0$ for all real $t$, so a quadratic has a nonpositive discriminant.
It implies the [[triangle inequality|0FJZMBX0ZZ]] for the induced norm.

## Explanations

### L1 human

The inner product of two vectors is at most the product of their lengths, in absolute value.
In the plane, this is the fact that $\abs{\cos\theta} \le 1$.
Equality happens only when the vectors lie on one line.