Cauchy–Schwarz inequality

Statement

Let $V$ be an inner product space over $\R$ or $\C$, with the induced norm $\norm{\cdot}$. 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 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.