Cauchy–Schwarz Inequality

|\langle x,y\rangle| \le \|x\|\,\|y\|

— the inner product never exceeds the product of the norms. Proved from the discriminant of

0 \le \|x-ty\|^2

; equality holds exactly for parallel vectors.

Examples & Exercises

Prove the Cauchy–Schwarz inequality $|\langle x,y\rangle| \le \|x\|\,\|y\|$ in $\mathbb{R}^n$ by the discriminant argument.

0 \le \|x-ty\|^2 = \|y\|^2 t^2 - 2\langle x,y\rangle t + \|x\|^2 \qquad (t \in \mathbb{R})

\Delta = 4\langle x,y\rangle^2 - 4\|x\|^2\|y\|^2 \le 0

\theta = 0

y = (1,0) \parallel x

, so

|\langle x,y\rangle| = 3 = \|x\|\,\|y\|

— the equality case (discriminant

0

) is visible.

Show proof

If $y = 0$ , both sides are $0$ and the claim is trivial. Assume $y \ne 0$ .
For every real $t$ , $0 \le \|x - ty\|^2$ . Expanding the inner product, $\|x - ty\|^2 = \langle x - ty, x - ty\rangle = \|y\|^2 t^2 - 2\langle x,y\rangle t + \|x\|^2$ , so the quadratic $\|y\|^2 t^2 - 2\langle x,y\rangle t + \|x\|^2$ in $t$ is always nonnegative.
A quadratic that is always nonnegative has nonpositive discriminant: $\Delta = 4\langle x,y\rangle^2 - 4\|y\|^2\|x\|^2 \le 0$ . Rearranging gives $\langle x,y\rangle^2 \le \|x\|^2\|y\|^2$ , i.e. $|\langle x,y\rangle| \le \|x\|\,\|y\|$ .
The same argument works in every inner product space. In $\ell^2$ , applying the $\mathbb{R}^N$ case to the first $N$ terms gives $|\sum_{n\le N} x_n y_n| \le \|x\|\,\|y\|$ ; letting $N \to \infty$ shows the series $\sum x_n y_n$ converges absolutely, so the inner product $\langle x,y\rangle = \sum_{n=1}^{\infty} x_n y_n$ is well defined.

Prove that equality $|\langle x,y\rangle| = \|x\|\,\|y\|$ in Cauchy–Schwarz holds if and only if $x$ and $y$ are linearly dependent.

|\langle x,y\rangle| = \|x\|\,\|y\| \iff x,\ y \text{ are linearly dependent}

\theta = \pi/2

x \perp y

, so

\langle x,y\rangle = 0 < 3 = \|x\|\,\|y\|

—

x

and

y

are linearly independent and equality fails.

Show proof

If $y = 0$ , both sides are $0$ so equality holds, and $\{x, 0\}$ is linearly dependent, making both directions trivial. Assume henceforth $y \ne 0$ .
( $\Leftarrow$ ) If $x$ and $y$ are linearly dependent, $x = cy$ for some scalar $c$ . Then $|\langle x,y\rangle| = |\langle cy,y\rangle| = |c|\,\|y\|^2 = \|cy\|\,\|y\| = \|x\|\,\|y\|$ .
( $\Rightarrow$ ) If equality holds, the discriminant of the quadratic $\|y\|^2 t^2 - 2\langle x,y\rangle t + \|x\|^2$ from the discriminant argument is $0$ , so it has a real root $t_0$ . At that root $0 = \|y\|^2 t_0^2 - 2\langle x,y\rangle t_0 + \|x\|^2 = \|x - t_0 y\|^2$ , hence $x = t_0 y$ — i.e. $x$ and $y$ are linearly dependent.

Deduce the triangle inequality $\|x+y\| \le \|x\| + \|y\|$ as a corollary of Cauchy–Schwarz.

\|x+y\|^2 = \|x\|^2 + 2\langle x,y\rangle + \|y\|^2 \le \|x\|^2 + 2\|x\|\,\|y\| + \|y\|^2 = (\|x\| + \|y\|)^2

\theta = \pi/4

\|x+y\| \approx 3.7739 < 4 = \|x\| + \|y\|

— the triangle inequality in action.

Show proof

Expanding the squared norm via the inner product gives $\|x+y\|^2 = \langle x+y, x+y\rangle = \|x\|^2 + 2\langle x,y\rangle + \|y\|^2$ .
Since $\langle x,y\rangle \le |\langle x,y\rangle|$ and $|\langle x,y\rangle| \le \|x\|\,\|y\|$ by Cauchy–Schwarz, we get $\|x+y\|^2 \le \|x\|^2 + 2\|x\|\,\|y\| + \|y\|^2 = (\|x\| + \|y\|)^2$ .
Taking square roots of both sides yields $\|x+y\| \le \|x\| + \|y\|$ .