Cauchy-Schwarz

( a · b)^2 ≤ | a|^2 | b|^2 — the universal vector inequality.

The idea

Theorem (Cauchy–Schwarz inequality).

For real vectors $\mathbf{a} = (a_1, \dots, a_n)$ and $\mathbf{b} = (b_1, \dots, b_n)$, $\left(\sum_i a_i b_i\right)^{2} \;\leq\; \left(\sum_i a_i^{2}\right)\left(\sum_i b_i^{2}\right),$ with equality if and only if one of the vectors is a scalar multiple of the other.

The sum $\sum_i a_i b_i$ is the dot product $\mathbf{a} \cdot \mathbf{b}$, and $\sum_i a_i^{2}$ is the squared length $\|\mathbf{a}\|^{2}$, so the inequality can be written $(\mathbf{a} \cdot \mathbf{b})^{2} \;\leq\; \|\mathbf{a}\|^{2}\,\|\mathbf{b}\|^{2}.$ In this form it bounds the dot product by the lengths alone: whatever the angle $\theta$ between the two vectors, $|\mathbf{a} \cdot \mathbf{b}|$ is at most $\|\mathbf{a}\|\,\|\mathbf{b}\|$. The bound is attained exactly when the vectors are parallel, which is the proportionality in the equality condition.

Cauchy–Schwarz is the standard tool for bounding a sum of products by two sums of squares, one measuring $\mathbf{a}$ alone and one measuring $\mathbf{b}$ alone.

Ways to work on it

• Walkthrough. Verify on a 2D pair, then identify the equality condition.
• Proof. Prove Cauchy–Schwarz in 2D via the Lagrange identity.
• Practice. Compute the left-hand side of the inequality for random 2D vectors.
• Hardest. Use Cauchy–Schwarz to maximize a linear form on the unit circle.

Not sure where to start? Take the ten-question placement test.