# Cauchy-Schwarz Inequality

The Cauchy-Schwarz Inequality (which is known by other names, including Cauchy's Inequality, Schwarz's Inequality, and the Cauchy-Bunyakovsky-Schwarz Inequality) is a well-known inequality with many elegant applications.

## Elementary Form

For any real numbers $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$,

$\left( \sum_{i=1}^{n}a_ib_i \right)^2 \le \left (\sum_{i=1}^{n}a_i^2 \right )\left (\sum_{i=1}^{n}b_i^2 \right )$,

with equality when there exist constants $\mu, \lambda$ not both zero such that for all $1 \le i \le n$, $\mu a_i = \lambda b_i$.

### Proof

There are several proofs; we will present an elegant one that does not generalize.

Consider the vectors $\mathbf{a} = \langle a_1, \ldots a_n \rangle$ and ${} \mathbf{b} = \langle b_1, \ldots b_n \rangle$. If $\theta$ is the angle formed by $\mathbf{a}$ and $\mathbf{b}$, then the left-hand side of the inequality is equal to the square of the dot product of $\mathbf{a}$ and $\mathbf{b}$, or $\left( ||\mathbf{a}|| \cdot ||\mathbf{b}|| \cos\theta \right)^2$. The right hand side of the inequality is equal to $\left( ||\mathbf{a}|| \cdot ||\mathbf{b}|| \right)^2$. The inequality then follows from $|\cos\theta | \le 1$, with equality when one of $\mathbf{a,b}$ is a multiple of the other, as desired.

### Complex Form

The inequality sometimes appears in the following form.

Let $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ be complex numbers. Then

$\left| \sum_{i=1}^na_ib_i \right|^2 \le \left (\sum_{i=1}^{n}|a_i^2|\right ) \left (\sum_{i=1}^n |b_i^2|\right )$.

This appears to be more powerful, but it follows immediately from

$\left| \sum_{i=1}^n a_ib_i \right| ^2 \le \left( \sum_{i=1}^n |a_i| \cdot |b_i| \right)^2 \le \left(\sum_{i=1}^n |a_i^2|\right)\left( \sum_{i=1}^n |b_i^2|\right )$.

## General Form

Let $V$ be a vector space, and let $\langle \cdot, \cdot \rangle : V \times V \mapsto \mathbb{R}$ be an inner product. Then for any $\mathbf{a,b} \in V$,

$\langle \mathbf{a,b} \rangle^2 \le \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle$,

with equality if and only if there exist constants $\mu, \lambda$ not both zero such that $\mu\mathbf{a} = \lambda\mathbf{b}$.

### Proof 1

Consider the polynomial of $t$

$\langle t\mathbf{a + b}, t\mathbf{a + b} \rangle = t^2\langle \mathbf{a,a} \rangle + 2t\langle \mathbf{a,b} \rangle + \langle \mathbf{b,b} \rangle$.

This must always be greater than or equal to zero, so it must have a non-positive discriminant, i.e., $\langle \mathbf{a,b} \rangle^2$ must be less than or equal to $\langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle$, with equality when $\mathbf{a = 0}$ or when there exists some scalar $-t$ such that $-t\mathbf{a} = \mathbf{b}$, as desired.

### Proof 2

We consider

$\langle \mathbf{a-b, a-b} \rangle = \langle \mathbf{a,a} \rangle + \langle \mathbf{b,b} \rangle - 2 \langle \mathbf{a,b} \rangle$.

Since this is always greater than or equal to zero, we have

$\langle \mathbf{a,b} \rangle \le \frac{1}{2} \langle \mathbf{a,a} \rangle + \frac{1}{2} \langle \mathbf{b,b} \rangle$.

Now, if either $\mathbf{a}$ or $\mathbf{b}$ is equal to $\mathbf{0}$, then $\langle \mathbf{a,b} \rangle^2 = \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle = 0$. Otherwise, we may normalize so that $\langle \mathbf {a,a} \rangle = \langle \mathbf{b,b} \rangle = 1$, and we have

$\langle \mathbf{a,b} \rangle \le 1 = \langle \mathbf{a,a} \rangle^{1/2} \langle \mathbf{b,b} \rangle^{1/2}$,

with equality when $\mathbf{a}$ and $\mathbf{b}$ may be scaled to each other, as desired.

### Examples

The elementary form of the Cauchy-Schwarz inequality is a special case of the general form, as is the Cauchy-Schwarz Inequality for Integrals: for integrable functions $f,g : [a,b] \mapsto \mathbb{R}$,

$\left( \int_{a}^b f(x)g(x)dx \right)^2 \le \int_{a}^b [f(x)]^2dx \cdot \int_a^b [g(x)]^2 dx$,

with equality when there exist constants $\mu, \lambda$ not both equal to zero such that for $t \in [a,b]$,

$\mu \int_a^t f(x)dx = \lambda \int_a^t g(x)dx$.

## Problems

### Introductory

• Consider the function $f(x)=\frac{(x+k)^2}{x^2+1},x\in (-\infty,\infty)$, where $k$ is a positive integer. Show that $f(x)\le k^2+1$.Source

### Intermediate

$P$ is a point inside a given triangle $ABC$. $D, E, F$ are the feet of the perpendiculars from $P$ to the lines $BC, CA, AB$, respectively. Find all $P$ for which
$\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}$