Cauchy-Schwarz Inequality

In algebra, the Cauchy-Schwarz Inequality, also known as the Cauchy–Bunyakovsky–Schwarz Inequality or informally as Cauchy-Schwarz, is an inequality with many ubiquitous formulations in abstract algebra, calculus, and contest mathematics. In high-school competitions, its applications are limited to elementary and linear algebra.

Its elementary algebraic formulation is often referred to as Cauchy's Inequality and states that for any list of reals $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ , $(a_1^2 + a_2^2 + \cdots + a_n^2)(b_1^2 + b_2^2 + \cdots + b_n^2) \geq (a_1b_1 + a_2b_2 + \cdots + a_nb_n)^2,$ with equality if and only if there exists a constant $t$ such that $a_n = t b_n$ for all $1 \leq t \leq n$ , or if every number in one of the lists is zero. Along with the AM-GM Inequality, Cauchy-Schwarz forms the foundation for inequality problems in intermediate and olympiad competitions. It is particularly crucial in proof-based contests.

Its linear algebraic forulation states that for any vectors $\overrightarrow{v}$ and $\overrightarrow{w}$ in $\mathbb{R}^n$ , $(\overrightarrow{v} \cdot \overrightarrow{v})(\overrightarrow{w} \cdot \overrightarrow{w}) \geq |\overrightarrow{v} \cdot \overrightarrow{w}|^2, \textrm{ or equivalently}, \textrm{ } \|\overrightarrow{v}\| \|\overrightarrow{w}\| \geq |\overrightarrow{v} \cdot \overrightarrow{w}|$ with equality if and only if there exists a scalar $t$ such that $\overrightarrow{v} = t \overrightarrow{w}$ , or if one of the vectors is zero. This formulation comes in handy in linear algebra problems at intermediate and olympiad problems.

The full Cauchy-Schwarz Inequality is written in terms of abstract vector spaces. Under this formulation, the elementary algebraic, linear algebraic, and calculus formulations are different cases of the general inequality.

Proof

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 $(\mathbf{a} \cdot \mathbf{b})^2 = a^2 b^2 (\cos\theta) ^2$ .The right hand side of the inequality is equal to $\left( ||\mathbf{a}|| * ||\mathbf{b}|| \right)^2 = a^2b^2$ . 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.

Lemmas

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

Upper Bound on (Σa)(Σb)

Let $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ be two sequences of positive real numbers with $0 < m \le \frac{a_i}{b_i} \le M$ for $1 \le i \le n$ . Then $\left(\sum_{i=1}^{n}a_i^2 \right) \left(\sum_{i=1}^{n}b_i^2 \right) \le \frac{(M+m)^2}{4Mm} \left( \sum_{i=1}^{n}a_ib_i \right)^2,$ with equality if and only if, for some ordering of the pairs $(a_i,b_i) \mapsto (a_{\sigma(i)},b_{\sigma(i)})$ , some $0 \le j \le n$ exists such that $a_{\sigma(i)}=mb_{\sigma(i)}$ for $1 \le \sigma(i) \le j$ and $a_{\sigma(i)}=Mb_{\sigma(i)}$ for $j+1 \le \sigma(i) \le n$ , and $m\sum_{\sigma(i)=1}^{j}b_{\sigma(i)}^2 = M\sum_{\sigma(i)=j+1}^{n}b_{\sigma(i)}^2.$ If we restrict that $m_1 \le a_i \le M_1$ and $m_2 \le b_i \le M_2$ for all $i$ , then it's clear that for $a_i/b_i$ to be $m=m_1/M_2$ or $M=M_1/m_2$ for all $i$ , then $a_i=m_1 \Longleftrightarrow b_i=M_2$ and $a_i=M_1 \Longleftrightarrow b_i=m_2$ , so $m\sum_{\sigma(i)=1}^{j}b_{\sigma(i)}^2 = M\sum_{\sigma(i)=j+1}^{n}b_{\sigma(i)}^2$ is equivalent to $\begin{align*} m(jM_2^2) = M((n-j)m_2^2) &\Longleftrightarrow m_1M_2j = M_1m_2(n-j)\\ &\Longleftrightarrow j = \left(\frac{M_1m_2}{M_1m_2+m_1M_2}\right) n. \end{align*}$ (When this is not an integer, the maximum occurs when $j$ is either the ceiling or floor of the right-hand side.) In the special case that $a_ib_i = k > 0$ is constant for all $i$ , we have $M_1=1/m_2$ and $m_1=1/M_2$ , so here $j$ must be $n/2$ .

Proof

Note that for all $i$ , we have $0 \le \left(\frac{a_i}{b_i}-m\right)\left(M-\frac{a_i}{b_i}\right) = \frac{1}{b_i^2}(a_ib_iM-a_i^2-b_i^2Mm+a_ib_im)$ or $(M+m)a_ib_i \ge a_i^2+(Mm)b_i^2,$ with equality if and only if $a_i=mb_i$ or $a_i=Mb_i$ . Summing up these inequalities over $1 \le i \le n$ , we obtain from AM-GM that $\begin{align*} (M+m)\sum_{i=1}^{n}a_ib_i &\ge \sum_{i=1}^{n}a_i^2 + (Mm)\sum_{i=1}^{n}b_i^2\\ &\ge 2\sqrt{Mm \left(\sum_{i=1}^{n}a_i^2 \right) \left(\sum_{i=1}^{n}b_i^2 \right)}, \end{align*}$ and squaring gives us the desired bound. For equality to occur, we must have $a_i=mb_i$ or $a_i=Mb_i$ for all $i$ . If, without loss of generality, $a_i=mb_i$ for $1 \le i \le j$ and $a_i=Mb_i$ for $j+1 \le i \le n$ for some $0 \le j \le n$ , then for the AM-GM to reach equality we must have (assume $M>m$ since $M=m$ is trivial) $\begin{align*} \sum_{i=1}^{n}a_i^2 &= Mm\sum_{i=1}^{n}b_i^2\\ m^2\sum_{i=1}^{j}b_i^2 + M^2\sum_{i=j+1}^{n}b_i^2 &= Mm\sum_{i=1}^{j}b_i^2 + Mm\sum_{i=j+1}^{n}b_i^2\\ (m-M)m\sum_{i=1}^{j}b_i^2 &= (m-M)M\sum_{i=j+1}^{n}b_i^2\\ m\sum_{i=1}^{j}b_i^2 &= M\sum_{i=j+1}^{n}b_i^2. \end{align*}$

General Form

Let $V$ be a vector space, and let $\langle \cdot, \cdot \rangle : V \times V \to \mathbb{R}$ 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.

Proof 3

Consider $a-\lambda b$ for some scalar $\lambda$ . Then: $0\le||a-\lambda b||^2$ (by the Trivial Inequality) $=\langle a-\lambda b,a-\lambda b\rangle$ $=\langle a,a\rangle-2\lambda\langle a,b\rangle+\lambda^2\langle y,y\rangle$ $=||a||^2-2\lambda\langle a,b\rangle+\lambda^2||b||^2$ . Now, let $\lambda=\frac{\langle a,b\rangle}{||b||^2}$ . Then, we have: $0\le||a||^2-\frac{\langle a,b\rangle|^2}{||b||^2}$ $\implies\langle a,b\rangle|^2\le||a||^2||b||^2=\langle a,a\rangle\cdot\langle b,b\rangle$ . $\square$

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 $f,g : [a,b] \mapsto \mathbb{R}$ , $\biggl( \int_{a}^b f(x)g(x)dx \biggr)^2 \le \int_{a}^b \bigl[ f(x) \bigr]^2dx \cdot \int_a^b \bigl[ g(x) \bigr]^2 dx$ 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)
(APMO 1991 #3) Let $a_1$ , $a_2$ , $\cdots$ , $a_n$ , $b_1$ , $b_2$ , $\cdots$ , $b_n$ be positive real numbers such that $a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n$ . Show that

$\frac {a_1^2}{a_1 + b_1} + \frac {a_2^2}{a_2 + b_2} + \cdots + \frac {a_n^2}{a_n + b_n} \geq \frac {a_1 + a_2 + \cdots + a_n}{2}$

Intermediate

Let $ABC$ be a triangle such that

$\left( \cot \frac{A}{2} \right)^2 + \left( 2 \cot \frac{B}{2} \right)^2 + \left( 3 \cot \frac{C}{2} \right)^2 = \left( \frac{6s}{7r} \right)^2 ,$ where $s$ and $r$ denote its semiperimeter and inradius, respectively. Prove that triangle $ABC$ is similar to a triangle $T$ whose side lengths are all positive integers with no common divisor and determine those integers. (Source)

Olympiad

$P$ 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