The Cauchy-Schwarz Inequality

Why This Matters

The Cauchy-Schwarz inequality is the bilinear backbone of olympiad inequalities. Its statement:

For real numbers $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ , $\left(\sum_{i=1}^n a_i b_i\right)^2 \le \left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right),$ with equality if and only if $(a_1, \ldots, a_n)$ and $(b_1, \ldots, b_n)$ are proportional (or one is the zero vector).

Cauchy-Schwarz is to bilinear sums what AM-GM is to additive and multiplicative structure: a fundamental lower bound, true in great generality, with one-line proofs in dozens of forms. Every olympiad student should know the statement, the equality case, at least one proof, and the Engel form (Titu's lemma) $\sum \frac{a_i^2}{b_i} \ge \frac{(\sum a_i)^2}{\sum b_i}$ , which is the form most contest problems actually need.

Recognize Cauchy-Schwarz when:

You have a sum of products and want to bound it by a product of sums (or the reverse).
You have ratios with squared numerators or denominators ( $\sum \frac{a_i^2}{b_i}$ or $\sum a_i^2 \sum \frac{1}{a_i^2}$ ).
A vector or inner-product geometry interpretation simplifies the problem.
The equality case is "vectors proportional", which is a strong cue.

The inequality lives in any inner-product space. The finite-sum form above is the discrete case; integral and $L^2$ -norm forms generalize it. Olympiad applications are almost entirely the discrete case.

The Inequality

Theorem

Cauchy-Schwarz Inequality (1821 / 1859 / 1885)

Statement

$\left(\sum_{i=1}^n a_i b_i\right)^2 \le \left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right)$ , with equality iff there exist scalars $(\lambda, \mu)$ not both zero with $\lambda a_i = \mu b_i$ for all $i$ .

Intuition

Geometric reading: $\left(\sum a_i b_i\right)^2 = \langle \mathbf{a}, \mathbf{b} \rangle^2 \le \|\mathbf{a}\|^2 \|\mathbf{b}\|^2$ . The angle $\theta$ between $\mathbf{a}$ and $\mathbf{b}$ satisfies $\cos^2 \theta \le 1$ , with equality iff $\theta = 0$ or $\pi$ , i.e., the vectors are parallel.

The inequality is exactly $|\cos \theta| \le 1$ for the angle between two vectors. That is the whole content.

Proof Sketch

Discriminant proof. Consider the quadratic in $t$ : $f(t) = \sum_{i=1}^n (a_i + t b_i)^2 = t^2 \sum b_i^2 + 2 t \sum a_i b_i + \sum a_i^2.$ This is non-negative for all real $t$ (sum of squares). Hence its discriminant is non-positive: $4 \left(\sum a_i b_i\right)^2 - 4 \left(\sum a_i^2\right) \left(\sum b_i^2\right) \le 0,$ which is Cauchy-Schwarz. Equality iff $f(t) = 0$ for some $t$ , i.e., $a_i = -t b_i$ for all $i$ , i.e., proportional.

Lagrange identity. Direct algebra: $\left(\sum a_i^2\right) \left(\sum b_i^2\right) - \left(\sum a_i b_i\right)^2 = \sum_{i < j} (a_i b_j - a_j b_i)^2.$ The right side is a sum of squares, hence non-negative. Equality iff every $a_i b_j - a_j b_i = 0$ , i.e., proportional.

Engel form / Titu's lemma. For positive $b_i$ , $\sum_{i=1}^n \frac{a_i^2}{b_i} \ge \frac{(\sum_{i=1}^n a_i)^2}{\sum_{i=1}^n b_i}.$ This rearranges to the standard form by setting $\tilde{a}_i = \frac{a_i}{\sqrt{b_i}}$ , $\tilde{b}_i = \sqrt{b_i}$ and expanding.

Why It Matters

Cauchy-Schwarz is the most-cited inequality in olympiad solutions after AM-GM. The Engel form (Titu's lemma) is especially powerful: it converts a sum-of-fractions inequality into a comparison of single ratios, which is often immediate from the problem hypothesis.

The inequality also generalizes in many directions: Holder's inequality (with exponents $p, q$ summing reciprocally to 1 instead of just $p = q = 2$ ), the Cauchy-Schwarz inequality on inner-product spaces, the integral form $\left(\int f g\right)^2 \le \int f^2 \int g^2$ , and the matrix-trace form. The same proof technique (discriminant, expanding a square, Lagrange identity) adapts to all of these.

Failure Mode

The inequality $\left(\sum a_i b_i\right)^2 \le \left(\sum a_i^2\right) \left(\sum b_i^2\right)$ is square. Without the square, you cannot conclude $\sum a_i b_i \le \sqrt{\sum a_i^2} \sqrt{\sum b_i^2}$ unless $\sum a_i b_i \ge 0$ . For sums of products with negative terms, the inequality bounds $|\sum a_i b_i|$ , not $\sum a_i b_i$ itself.

The Engel form $\sum \frac{a_i^2}{b_i} \ge \frac{(\sum a_i)^2}{\sum b_i}$ requires $b_i > 0$ . If any $b_i = 0$ the fraction is undefined. If any $b_i < 0$ the inequality direction can flip.

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Worked Example: The Engel Form in Action

Example

The Engel form (Titu's lemma) at n=3

Problem. For positive reals $a, b, c, x, y, z$ , show $\frac{a^2}{x} + \frac{b^2}{y} + \frac{c^2}{z} \ge \frac{(a + b + c)^2}{x + y + z}.$

Solution. This is exactly the Engel form (Titu's lemma) at $n = 3$ . Apply Cauchy-Schwarz to the sequences $\left(\frac{a}{\sqrt{x}}, \frac{b}{\sqrt{y}}, \frac{c}{\sqrt{z}}\right)$ and $\left(\sqrt{x}, \sqrt{y}, \sqrt{z}\right)$ : $\left(\frac{a}{\sqrt{x}} \cdot \sqrt{x} + \frac{b}{\sqrt{y}} \cdot \sqrt{y} + \frac{c}{\sqrt{z}} \cdot \sqrt{z}\right)^2 \le \left(\frac{a^2}{x} + \frac{b^2}{y} + \frac{c^2}{z}\right) (x + y + z).$ The left side is $(a + b + c)^2$ . Dividing by $x + y + z$ gives the result.

Equality iff $\frac{a/\sqrt{x}}{\sqrt{x}} = \frac{b/\sqrt{y}}{\sqrt{y}} = \frac{c/\sqrt{z}}{\sqrt{z}}$ , i.e., $\frac{a}{x} = \frac{b}{y} = \frac{c}{z}$ .

The Engel form is what you actually use in 80% of contest problems. Recognizing the pattern $\sum \frac{(\text{thing})^2}{(\text{positive thing})}$ is the entry point.

Worked Example: A Classic Olympiad

Example

Nesbitt's inequality

Problem. For positive reals $a, b, c$ , show $\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} \ge \frac{3}{2}.$

Solution (via Engel). Rewrite each fraction as $\frac{a^2}{a(b + c)}$ : $\sum \frac{a}{b + c} = \sum \frac{a^2}{a(b + c)} \ge \frac{(a + b + c)^2}{\sum a(b + c)} = \frac{(a + b + c)^2}{2(ab + bc + ca)}.$

It remains to show $\frac{(a + b + c)^2}{2(ab + bc + ca)} \ge \frac{3}{2}$ , i.e., $(a + b + c)^2 \ge 3(ab + bc + ca)$ . Expand the left: $a^2 + b^2 + c^2 + 2(ab + bc + ca) \ge 3(ab + bc + ca)$ , i.e., $a^2 + b^2 + c^2 \ge ab + bc + ca$ , which is the classic SOS identity $\frac{1}{2}((a - b)^2 + (b - c)^2 + (c - a)^2) \ge 0$ .

Equality at $a = b = c$ .

Two-step pattern: Engel reduces to a polynomial inequality; SOS or direct expansion finishes.

The Recognition Process

When facing an inequality, ask:

Is there a sum of products and a product of sums? That is the standard CS form.
Is there a sum of fractions with squared numerators? The Engel form fits: $\sum \frac{a_i^2}{b_i} \ge \frac{(\sum a_i)^2}{\sum b_i}$ with $b_i > 0$ .
Is the equality case "vectors proportional"? A strong indicator that CS is the right tool.
Can you write each term as $a_i b_i$ for a clever choice of $a_i$ and $b_i$ ? The trick in many problems is the creative choice of which factors play the roles of $a$ and $b$ .
Should you use Holder instead? If the inequality is "more than bilinear" (cubic ratios, mixed exponents), Holder's $\left(\sum a_i^p\right)^{1/p} \left(\sum b_i^q\right)^{1/q} \ge \sum a_i b_i$ for $1/p + 1/q = 1$ is often the extension you need.

Common Mistakes

Watch Out

Forgetting the square

Cauchy-Schwarz is $\left(\sum a_i b_i\right)^2 \le \cdots$ , not $\sum a_i b_i \le \cdots$ . The square matters when the sum of products can be negative. Always check signs before removing the square.

Watch Out

Wrong choice of $a_i$ and $b_i$

The art of using CS is choosing the right pair of sequences. A generic choice usually gives a useless bound. Look at the LHS and RHS of what you want to prove; reverse-engineer what $a_i b_i$ products would produce the LHS and what $a_i^2$ and $b_i^2$ sums would produce the RHS.

Watch Out

Equality conditions glossed over

The equality case ( $a_i / b_i$ constant) is essential for optimization problems. Many writeups apply CS, get a bound, and forget to check whether equality is achievable in the problem's constraints. If equality cannot be achieved, the bound is not tight.

Watch Out

CS does not always give the tightest bound

For some inequalities, CS gives the right answer; for others, it gives a bound that is strictly weaker than the truth. If CS gives a bound that the problem hypothesis says cannot be achieved, you have not solved the problem. Either CS is the wrong tool or you need a sharper variant (Holder, rearrangement, SOS).

Watch Out

Engel form requires $b_i > 0$

The Engel form $\sum \frac{a_i^2}{b_i} \ge \frac{(\sum a_i)^2}{\sum b_i}$ needs every $b_i > 0$ . If a $b_i$ is zero or negative, this form fails (the LHS is undefined or the inequality flips). Verify positivity before applying.

Exercises

ExerciseCore

Problem

For real numbers $a, b, c$ , show $(a + b + c)^2 \le 3(a^2 + b^2 + c^2)$ .

ExerciseCore

Problem

For positive reals $a_1, \ldots, a_n$ , show $\sum a_i \cdot \sum \frac{1}{a_i} \ge n^2$ .

ExerciseAdvanced

Problem

Show that for positive reals $a, b, c$ with $a + b + c = 1$ , $\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} \ge \frac{9}{2}.$

Cross-Network Links

ProofsPath: am-gm-inequality is the multiplicative cousin; power-mean-inequality generalizes both; jensen-for-convex-functions is the convexity-based generalization; rearrangement-inequality handles cases CS does not.
TheoremPath: inner-product-spaces-and-orthogonality is the abstract setting where Cauchy-Schwarz lives in full generality; maximum-likelihood-estimation uses Cauchy-Schwarz in proofs of the Cramer-Rao lower bound; matrix-norms uses CS in the operator-norm submultiplicativity proof.

References

See structured references block. Primary entry points: Steele Cauchy-Schwarz Master Class for the canonical olympiad treatment with dozens of variations and exercises; Hardy-Littlewood-Polya Inequalities Ch 2 for the classical analytic perspective; Engel Ch 7 for contest applications.