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 is useful in linear algebra questions in contest mathematics

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.

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