# Cauchy-Schwarz Inequality

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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)

• $P$ 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.