Difference between revisions of "Cauchy-Schwarz Inequality"

Latest revision as of 14:28, 22 February 2024

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 $a_i = t b_i$ for all $1 \leq i \leq n$ , or if one list consists of only zeroes. 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 vector formulation states that for any vectors $\overrightarrow{v}$ and $\overrightarrow{w}$ in $\mathbb{R}^n$ , where $\overrightarrow{v} \cdot \overrightarrow{w}$ is the dot product of $\overrightarrow{v}$ and $\overrightarrow{w}$ and $\| \overrightarrow{v} \|$ is the norm of $\overrightarrow{v}$ , $\|\overrightarrow{v}\| \|\overrightarrow{w}\| \geq |\overrightarrow{v} \cdot \overrightarrow{w}|$ 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.

Proofs

Here is a list of proofs of Cauchy-Schwarz.

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

A Useful Inequality

Also known as Sedrakyan's Inequality, Bergström's Inequality, Engel's Form or Titu's Lemma the following inequality is a direct result of Cauchy-Schwarz inequality:

For any real numbers $(a_1,a_2,...,a_n)$ and $(b_1,b_2,...,b_n)$ where $(b_i>0, i\in \{1,2,..,n\})$ the following is true: $\frac{a_1^{2}}{b_1}+\frac{a_2^{2}}{b_2}+\ldots+\frac{a_n^{2}}{b_n}\geq\frac{\left(a_1+a_2+...+a_n\right)^{2}}{b_1+b_2+...+b_n}.$

Real Vector Spaces

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 $\mu\mathbf{a} = \lambda\mathbf{b}$ . The following proofs assume the inner product to be real-valued and commutative, and so only apply to vector spaces over the real numbers.

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: $0\le||a||^2-\frac{\langle a,b\rangle|^2}{||b||^2}$ $\implies\langle a,b\rangle|^2\le||a||^2||b||^2=\langle a,a\rangle\cdot\langle b,b\rangle$ . $\square$

Complex Vector Spaces

For any two vectors $\mathbf{a}, \mathbf{b}$ in the complex vector space $W$ , the following holds: $|\langle \mathbf{a}, \mathbf{b}\rangle| \leq ||\mathbf{a}||||\mathbf{b}||$ with equality holding only when $\mathbf{a}, \mathbf{b}$ are linearly dependent.

Proof

The following proof, a geometric argument that uses only the algebraic properties of the inner product, was discovered by Tarung Bhimnathwala in 2021.

Define the unit vectors $\mathbf{u}$ , $\mathbf{v}$ as $\mathbf{u} = \frac{\mathbf{a}}{||\mathbf{a}||}$ and $\mathbf{v} = \frac{\mathbf{b}}{||\mathbf{b}||}$ . Put $\gamma = \frac{\langle \mathbf{u},\mathbf{v}\rangle}{|\langle \mathbf{u},\mathbf{v}\rangle|}$ . In other words, $\gamma$ is the complex argument of $\langle \mathbf{u},\mathbf{v}\rangle$ and lies on the unit circle. If any of the denominators are zero, the entire result follows trivially. Let $\mathbf{p} = \frac{1}{2}(\mathbf{u}+\gamma \mathbf{v})$ and $\mathbf{q} = \frac{1}{2}(\mathbf{u}-\gamma \mathbf{v})$ . Importantly, we have $\langle \mathbf{p},\mathbf{q}\rangle = \frac{1}{4}(||\mathbf{u}||^2 -\langle \mathbf{u},\gamma \mathbf{v}\rangle + \langle \gamma \mathbf{v},\mathbf{u}\rangle - \gamma \bar{\gamma}||\mathbf{v}||^2)$ $= \frac{1}{4}(1 -\langle \mathbf{u},\gamma \mathbf{v}\rangle + \langle \gamma \mathbf{v},\mathbf{u}\rangle - 1)$ $= \frac{1}{4}(\langle \gamma \mathbf{v},\mathbf{u}\rangle - \overline{\langle \gamma \mathbf{v},\mathbf{u}\rangle})$ $= \frac{1}{2}\operatorname{Im}(\langle \gamma \mathbf{v},\mathbf{u}\rangle)$ $= \frac{1}{2}\operatorname{Im}(\gamma\overline{\langle \mathbf{u},\mathbf{v}\rangle})$ $= 0.$ Since $\mathbf{u}=\mathbf{p}+\mathbf{q}$ and $\mathbf{v} = \bar{\gamma}(\mathbf{p}-\mathbf{q})$ , this calculation shows that $\mathbf{p}$ and $\mathbf{q}$ form an orthogonal basis of the linear subspace spanned by $\mathbf{u}$ and $\mathbf{v}$ . Thus we can think of $\mathbf{u}$ and $\mathbf{v}$ as lying on the unit sphere in this subspace, which is isomorphic to $\mathbb{C}^2$ . Another thing to note is that $||\mathbf{p}||^2 + ||\mathbf{q}||^2 = \langle \mathbf{p},\mathbf{p} \rangle + \langle \mathbf{q},\mathbf{q} \rangle$ $= \frac{1}{4}(\langle \mathbf{u}+\gamma \mathbf{v},\mathbf{u}+\gamma \mathbf{v}\rangle + \langle \mathbf{u}-\gamma \mathbf{v},\mathbf{u}-\gamma \mathbf{v}\rangle)$ $= \frac{1}{4}(2\langle \mathbf{u},\mathbf{u} \rangle + 2\gamma\bar{\gamma}\langle \mathbf{v},\mathbf{v} \rangle)$ $= \frac{1}{4}(2||\mathbf{u}||^2 + 2||\mathbf{v}||^2)$ $= 1.$

The previous two calculations established that $\mathbf{p}$ and $\mathbf{q}$ are orthogonal, and that the sum of their squared norms is $1$ . Now we have $|\langle \mathbf{u},\mathbf{v}\rangle| = |\langle \mathbf{p}+\mathbf{q},\bar{\gamma}(\mathbf{p}-\mathbf{q})\rangle|$ $= |\gamma||\langle \mathbf{p}+\mathbf{q},\mathbf{p}-\mathbf{q}\rangle|$ $= |\langle \mathbf{p}+\mathbf{q},\mathbf{p}-\mathbf{q}\rangle|$ $= |||\mathbf{p}||^2 +\langle \mathbf{q},\mathbf{p}\rangle - \langle \mathbf{p},\mathbf{q}\rangle - ||\mathbf{q}||^2|$ $= |||\mathbf{p}||^2 - ||\mathbf{q}||^2|$ $= |||\mathbf{p}||^2 + ||\mathbf{q}||^2 - 2||\mathbf{q}||^2|$ $= |1 - 2||\mathbf{q}||^2|$ $\leq 1.$ Equality holds when either $||\mathbf{q}||=0$ or $||\mathbf{q}||=1$ , or equivalently when $\mathbf{u}=\pm \mathbf{v}$ and $\mathbf{a} = \lambda \mathbf{b}$ . Lastly, multiplying each side by $||\mathbf{a}||||\mathbf{b}||$ , we have $|\langle \mathbf{a},\mathbf{b}\rangle| \leq ||\mathbf{a}||||\mathbf{b}||.$

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

$\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}$ is least.

(Source)

Other Resources

Wikipedia entry

Books

The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities by J. Michael Steele.
Problem Solving Strategies by Arthur Engel contains significant material on inequalities.

@@ Line 1: / Line 1: @@
-The '''Cauchy-Schwarz Inequality''' (which is known by other names, including Cauchy's Inequality) states that, for two sets of real numbers <math>a_1,a_2,\ldots,a_n</math> and <math>b_1,b_2,\ldots,b_n</math>, the following [[inequality]] is always true:
+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 <math>a_1, a_2, \ldots, a_n</math> and <math>b_1, b_2, \ldots, b_n</math>, <cmath>(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,</cmath> with equality if and only if there exists a constant <math>t</math> such that <math>a_i = t b_i</math> for all <math>1 \leq i \leq n</math>, or if one list consists of only zeroes. 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.
-<math> \displaystyle ({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</math>
+Its vector formulation states that for any vectors <math>\overrightarrow{v}</math> and <math>\overrightarrow{w}</math> in <math>\mathbb{R}^n</math>, where <math>\overrightarrow{v} \cdot \overrightarrow{w}</math> is the [[dot product]] of <math>\overrightarrow{v}</math> and <math>\overrightarrow{w}</math> and <math>\| \overrightarrow{v} \|</math> is the [[norm]] of <math>\overrightarrow{v}</math>, <cmath>\|\overrightarrow{v}\| \|\overrightarrow{w}\| \geq |\overrightarrow{v} \cdot \overrightarrow{w}|</cmath> with equality if and only if there exists a scalar <math>t</math> such that <math>\overrightarrow{v} = t \overrightarrow{w}</math>, 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.
-[[Equality condition| equality]] holds if and only if <math>\frac{a_1}{b_1}=\frac{a_2}{b_2}= \cdots =\frac{a_n}{b_n}</math>.
+== Proofs ==
+Here is a list of proofs of Cauchy-Schwarz.
+Consider the vectors <math> \mathbf{a} = \langle a_1, \ldots a_n \rangle </math> and <math> {} \mathbf{b} = \langle b_1, \ldots b_n \rangle </math>.  If <math>\theta </math> is the [[angle]] formed by <math> \mathbf{a} </math> and <math> \mathbf{b} </math>, then the left-hand side of the inequality is equal to the square of the [[dot product]] of <math>\mathbf{a} </math> and <math> \mathbf{b} </math>, or <math>(\mathbf{a} \cdot \mathbf{b})^2 = a^2 b^2 (\cos\theta) ^2</math>  .The right hand side of the inequality is equal to <math> \left( ||\mathbf{a}|| * ||\mathbf{b}|| \right)^2  =  a^2b^2</math>.  The inequality then follows from <math> |\cos\theta | \le 1 </math>, with equality when one of <math> \mathbf{a,b} </math> is a multiple of the other, as desired.
-== Proof ==
+== Lemmas ==
-There are many ways to prove this; one of the more well-known is to consider the equation
-<math> \displaystyle (a_1x+b_1)^2+(a_2x+b_2)^2+ \cdots +(a_nx+b_n)^2=0</math>.
+=== Complex Form ===
+The inequality sometimes appears in the following form.
-Expanding, we find the equation to be of the form
+Let <math> a_1, \ldots, a_n </math> and <math> b_1, \ldots, b_n </math> be [[complex numbers]].  Then
+<cmath>
+\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)
+</cmath>
+This appears to be more powerful, but it follows from
+<cmath>
+\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)
+</cmath>
-<math> \displaystyle Ax^2 + Bx + C,</math>
+=== A Useful Inequality ===
+Also known as Sedrakyan's Inequality,  Bergström's Inequality, Engel's Form or Titu's Lemma the following inequality is a direct result of Cauchy-Schwarz inequality:
-where <math>A=\sum_{i=1}^n a_i^2</math>, <math>B=2\sum_{j=1}^n a_jb_j</math>, and <math>C=\sum_{k=1}^n b_k^2.</math>.  By the [[Trivial inequality | Trivial Inequality]], we know that the left-hand-side of the original equation is always at least 0, so either both roots are [[complex numbers]], or there is a double root at <math>x=0</math>.  Either way, the [[discriminant]] of the equation is nonpositive.  Taking the [[discriminant]], <math>B^2-4AC \leq 0</math> and substituting the above values of A, B, and C leaves us with the Cauchy-Schwarz Inequality,
+For any real numbers <math>(a_1,a_2,...,a_n)</math> and <math>(b_1,b_2,...,b_n)</math> where <math>(b_i>0, i\in \{1,2,..,n\})</math> the following is true: <cmath> \frac{a_1^{2}}{b_1}+\frac{a_2^{2}}{b_2}+\ldots+\frac{a_n^{2}}{b_n}\geq\frac{\left(a_1+a_2+...+a_n\right)^{2}}{b_1+b_2+...+b_n}.</cmath>
+== Real Vector Spaces ==
-<math> \displaystyle (a_1b_1+a_2b_2+ \cdots +a_nb_n)^2 \leq (a_1^2+a_2^2+ \cdots +a_n^2)(b_1^2+b_2^2+ \cdots +b_n^2),</math>
+Let <math>V </math> be a [[vector space]], and let <math> \langle \cdot, \cdot \rangle : V \times V \to \mathbb{R} </math> be an [[inner product]].  Then for any <math> \mathbf{a,b} \in V </math>,
+<cmath>
+\langle \mathbf{a,b} \rangle^2 \le \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle ,
+</cmath>
+with equality if and only if there exist constants <math>\mu, \lambda </math> not both zero such that <math> \mu\mathbf{a} = \lambda\mathbf{b} </math>. The following proofs assume the inner product to be real-valued and commutative, and so only apply to vector spaces over the real numbers.
+=== Proof 1 ===
-or, in the more compact [[sigma notation]],
+Consider the polynomial of <math> t </math>
+<cmath>
+\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 .
+</cmath>
+This must always be greater than or equal to zero, so it must have a non-positive discriminant, i.e., <math> \langle \mathbf{a,b} \rangle^2 </math> must be less than or equal to <math> \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle </math>, with equality when <math> \mathbf{a = 0} </math> or when there exists some scalar <math>-t </math> such that <math> -t\mathbf{a} = \mathbf{b} </math>, as desired.
-<math>{\left(\sum a_ib_i\right)}^2 \leq \left(\sum a_i^2\right)\left(\sum b_i^2\right).</math>
+=== Proof 2 ===
+We consider
-Note that this also gives us the equality case; equality holds if and only if the discriminant is equal to 0, which is true if and only if the equation has 0 as a double root, which is true if and only if <math>\frac{a_1}{b_1}=\frac{a_2}{b_2}= \cdots =\frac{a_n}{b_n}</math>.
+<cmath>
+\langle \mathbf{a-b, a-b} \rangle = \langle \mathbf{a,a} \rangle + \langle \mathbf{b,b} \rangle - 2 \langle \mathbf{a,b} \rangle .
+</cmath>
+Since this is always greater than or equal to zero, we have
+<cmath>
+\langle \mathbf{a,b} \rangle \le \frac{1}{2} \langle \mathbf{a,a} \rangle + \frac{1}{2} \langle \mathbf{b,b} \rangle .
+</cmath>
+Now, if either <math> \mathbf{a} </math> or <math> \mathbf{b} </math> is equal to <math> \mathbf{0} </math>, then <math> \langle \mathbf{a,b} \rangle^2 = \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle = 0 </math>.  Otherwise, we may [[normalize]] so that <math> \langle \mathbf {a,a} \rangle = \langle \mathbf{b,b} \rangle = 1 </math>, and we have
+<cmath>
+\langle \mathbf{a,b} \rangle \le 1 = \langle \mathbf{a,a} \rangle^{1/2} \langle \mathbf{b,b} \rangle^{1/2} ,
+</cmath>
+with equality when <math>\mathbf{a} </math> and <math> \mathbf{b} </math> may be scaled to each other, as desired.
-== Contest Problem Solving ==
+=== Proof 3 ===
-This inequality is used very frequently to solve Olympiad-level Inequality problems, such as those on the [[United States of America Mathematics Olympiad | USAMO]] and [[International Mathematics Olympiad | IMO]].
+Consider <math>a-\lambda b</math> for some scalar <math>\lambda</math>. Then:
+<math>0\le||a-\lambda b||^2</math> (by the Trivial Inequality)
+<math>=\langle a-\lambda b,a-\lambda b\rangle</math>
+<math>=\langle a,a\rangle-2\lambda\langle a,b\rangle+\lambda^2\langle y,y\rangle</math>
+<math>=||a||^2-2\lambda\langle a,b\rangle+\lambda^2||b||^2</math>.
+Now, let <math>\lambda=\frac{\langle a,b\rangle}{||b||^2}</math>. Then, we have:
+<math>0\le||a||^2-\frac{\langle a,b\rangle|^2}{||b||^2}</math>
+<math>\implies\langle a,b\rangle|^2\le||a||^2||b||^2=\langle a,a\rangle\cdot\langle b,b\rangle</math>. <math>\square</math>
+== Complex Vector Spaces ==
+For any two vectors <math>\mathbf{a}, \mathbf{b}</math> in the complex vector space <math>W</math>, the following holds:
+<cmath>|\langle \mathbf{a}, \mathbf{b}\rangle| \leq ||\mathbf{a}||||\mathbf{b}||</cmath> with equality holding only when <math>\mathbf{a}, \mathbf{b}</math> are linearly dependent.
+=== Proof ===
+The following proof, a geometric argument that uses only the algebraic properties of the inner product, was discovered by Tarung Bhimnathwala in 2021.
+Define the unit vectors <math>\mathbf{u}</math>, <math>\mathbf{v}</math> as <math>\mathbf{u} = \frac{\mathbf{a}}{||\mathbf{a}||}</math> and <math>\mathbf{v} = \frac{\mathbf{b}}{||\mathbf{b}||}</math>. Put <math>\gamma = \frac{\langle \mathbf{u},\mathbf{v}\rangle}{|\langle \mathbf{u},\mathbf{v}\rangle|}</math>. In other words, <math>\gamma</math> is the complex argument of <math>\langle \mathbf{u},\mathbf{v}\rangle</math> and lies on the unit circle. If any of the denominators are zero, the entire result follows trivially. Let <math>\mathbf{p} = \frac{1}{2}(\mathbf{u}+\gamma \mathbf{v})</math> and <math>\mathbf{q} = \frac{1}{2}(\mathbf{u}-\gamma \mathbf{v})</math>. Importantly, we have
+<cmath>
+    \langle \mathbf{p},\mathbf{q}\rangle = \frac{1}{4}(||\mathbf{u}||^2 -\langle \mathbf{u},\gamma \mathbf{v}\rangle + \langle \gamma \mathbf{v},\mathbf{u}\rangle - \gamma \bar{\gamma}||\mathbf{v}||^2)
+</cmath>
+<cmath>
+    = \frac{1}{4}(1 -\langle \mathbf{u},\gamma \mathbf{v}\rangle + \langle \gamma \mathbf{v},\mathbf{u}\rangle - 1)
+</cmath>
+<cmath>
+    = \frac{1}{4}(\langle \gamma \mathbf{v},\mathbf{u}\rangle - \overline{\langle \gamma \mathbf{v},\mathbf{u}\rangle})
+</cmath>
+<cmath>
+    = \frac{1}{2}\operatorname{Im}(\langle \gamma \mathbf{v},\mathbf{u}\rangle)
+</cmath>
+<cmath>
+    = \frac{1}{2}\operatorname{Im}(\gamma\overline{\langle \mathbf{u},\mathbf{v}\rangle})
+</cmath>
+<cmath>
+    = 0.
+</cmath>
+Since <math>\mathbf{u}=\mathbf{p}+\mathbf{q}</math> and <math>\mathbf{v} = \bar{\gamma}(\mathbf{p}-\mathbf{q})</math>, this calculation shows that <math>\mathbf{p}</math> and <math>\mathbf{q}</math> form an orthogonal basis of the linear subspace spanned by <math>\mathbf{u}</math> and <math>\mathbf{v}</math>. Thus we can think of <math>\mathbf{u}</math> and <math>\mathbf{v}</math> as lying on the unit sphere in this subspace, which is isomorphic to <math>\mathbb{C}^2</math>. Another thing to note is that
+<cmath>
+    ||\mathbf{p}||^2 + ||\mathbf{q}||^2 = \langle \mathbf{p},\mathbf{p} \rangle + \langle \mathbf{q},\mathbf{q} \rangle
+</cmath>
+<cmath>
+    = \frac{1}{4}(\langle \mathbf{u}+\gamma \mathbf{v},\mathbf{u}+\gamma \mathbf{v}\rangle + \langle \mathbf{u}-\gamma \mathbf{v},\mathbf{u}-\gamma \mathbf{v}\rangle)
+</cmath>
+<cmath>
+    = \frac{1}{4}(2\langle \mathbf{u},\mathbf{u} \rangle + 2\gamma\bar{\gamma}\langle \mathbf{v},\mathbf{v} \rangle)
+</cmath>
+<cmath>
+    = \frac{1}{4}(2||\mathbf{u}||^2 + 2||\mathbf{v}||^2)
+</cmath>
+<cmath>
+    = 1.
+</cmath>
+The previous two calculations established that <math>\mathbf{p}</math> and <math>\mathbf{q}</math> are orthogonal, and that the sum of their squared norms is <math>1</math>. Now we have
+<cmath>
+    |\langle \mathbf{u},\mathbf{v}\rangle| = |\langle \mathbf{p}+\mathbf{q},\bar{\gamma}(\mathbf{p}-\mathbf{q})\rangle|
+</cmath>
+<cmath>
+    = |\gamma||\langle \mathbf{p}+\mathbf{q},\mathbf{p}-\mathbf{q}\rangle|
+</cmath>
+<cmath>
+    = |\langle \mathbf{p}+\mathbf{q},\mathbf{p}-\mathbf{q}\rangle|
+</cmath>
+<cmath>
+    = |||\mathbf{p}||^2 +\langle \mathbf{q},\mathbf{p}\rangle - \langle \mathbf{p},\mathbf{q}\rangle - ||\mathbf{q}||^2|
+</cmath>
+<cmath>
+    = |||\mathbf{p}||^2 - ||\mathbf{q}||^2|
+</cmath>
+<cmath>
+    = |||\mathbf{p}||^2 + ||\mathbf{q}||^2 - 2||\mathbf{q}||^2|
+</cmath>
+<cmath>
+    = |1 - 2||\mathbf{q}||^2|
+</cmath>
+<cmath>
+    \leq 1.
+</cmath>
+Equality holds when either <math>||\mathbf{q}||=0</math> or <math>||\mathbf{q}||=1</math>, or equivalently when <math>\mathbf{u}=\pm \mathbf{v}</math> and <math>\mathbf{a} = \lambda \mathbf{b}</math>. Lastly, multiplying each side by <math>||\mathbf{a}||||\mathbf{b}||</math>, we have <cmath>|\langle \mathbf{a},\mathbf{b}\rangle| \leq ||\mathbf{a}||||\mathbf{b}||.</cmath>
+==Problems==
+===Introductory===
+*Consider the function <math>f(x)=\frac{(x+k)^2}{x^2+1},x\in (-\infty,\infty)</math>, where <math>k</math> is a positive integer. Show that <math>f(x)\le k^2+1</math>. ([[User:Temperal/The_Problem_Solver's Resource Competition|Source]])
+* [http://www.mathlinks.ro/Forum/viewtopic.php?t=78687 (APMO 1991 #3)] Let <math>a_1</math>, <math>a_2</math>, <math>\cdots</math>, <math>a_n</math>, <math>b_1</math>, <math>b_2</math>, <math>\cdots</math>, <math>b_n</math> be positive real numbers such that <math>a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n</math>.  Show that
+<cmath>\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}</cmath>
+===Intermediate===
+*Let <math>ABC</math> be a triangle such that
+<cmath>
+\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 ,
+</cmath>
+where <math>s</math> and <math>r</math> denote its [[semiperimeter]] and [[inradius]], respectively.  Prove that triangle <math>ABC </math> is similar to a triangle <math>T </math> whose side lengths are all positive integers with no common divisor and determine those integers.
+([[2002 USAMO Problems/Problem 2|Source]])
+===Olympiad===
+*<math>P</math> is a point inside a given triangle <math>ABC</math>.  <math>D, E, F</math> are the feet of the perpendiculars from <math>P</math> to the lines <math>BC, CA, AB</math>, respectively.  Find all <math>P</math> for which
+<cmath>
+\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}
+</cmath>
+is least.
+([[1981 IMO Problems/Problem 1|Source]])
 == Other Resources ==
-* [http://en.wikipedia.org/wiki/Cauchy-Schwartz_inequality Wikipedia entry]
+* [http://en.wikipedia.org/wiki/Cauchy-Schwarz_inequality Wikipedia entry]
 ===Books===
 * [http://www.amazon.com/exec/obidos/ASIN/052154677X/artofproblems-20 The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities] by J. Michael Steele.
 * [http://www.amazon.com/exec/obidos/ASIN/0387982191/artofproblems-20 Problem Solving Strategies] by Arthur Engel contains significant material on inequalities.
+[[Category:Algebra]]
+[[Category:Inequalities]]

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