2019년 2월 7일 목요일

Cauchy-Schwarz Inequality 코시슈바르츠 부등식

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. It has an elementary form, a complex form, and a general form.
Louis Cauchy wrote the first paper about the elementary form in 1821. The general form was discovered by Bunyakovsky in 1849 and independently by Schwarz in 1888.

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 exists a nonzero constant $\mu$ such that for all $1 \le i \le n$, $\mu a_i = b_i$.

Discussion

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.

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




AoPS
작성자: 시간:

댓글 없음:

댓글 쓰기

피드 구독하기: 댓글 (Atom)