Difference between revisions of "Cauchy-Schwarz Inequality"

Latest revision as of 14:57, 17 January 2021

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.

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

$\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}$ 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''', '''Schwarz's Inequality''', and the '''Cauchy-Bunyakovsky-Schwarz Inequality''') is a well-known [[inequality]] with many elegant applications.
+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.
+[https://en.wikipedia.org/wiki/Augustin-Louis_Cauchy|Augustin Louis Cauchy] wrote the first paper about the elementary form in 1821. The general form was discovered by [https://en.wikipedia.org/wiki/Viktor_Bunyakovsky|Viktor Bunyakovsky] in 1849 and independently by [https://en.wikipedia.org/wiki/Hermann_Amandus_Schwarz|Hermann Schwarz] in 1888.
 == Elementary Form ==
 For any real numbers <math> a_1, \ldots, a_n </math> and <math> b_1, \ldots, b_n </math>,
-<center>
+<cmath>
-<math>
+\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)
-\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 )
+</cmath>
-</math>,
+with equality when there exists a nonzero constant <math>\mu</math> such that for all <math> 1 \le i \le n </math>, <math>\mu a_i = b_i </math>.
-</center>
-with equality when there exist constants <math>\mu, \lambda </math> not both zero such that for all <math> 1 \le i \le n </math>, <math>\mu a_i = \lambda b_i </math>.
-=== Proof ===
+=== Discussion ===
-There are several proofs; we will present an elegant one that does not generalize.
+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.
-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> \left( ||\mathbf{a}|| \cdot ||\mathbf{b}|| \cos\theta \right)^2 </math>.  The right hand side of the inequality is equal to <math> \left( ||\mathbf{a}|| \cdot ||\mathbf{b}|| \right)^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. Note that this is not actually a proof; rather, this result is used to establish that we can define the angles between two vectors in this way.
 === Complex Form ===
@@ Line 22: / Line 20: @@
 Let <math> a_1, \ldots, a_n </math> and <math> b_1, \ldots, b_n </math> be [[complex numbers]].  Then
-<center>
+<cmath>
-<math>
+\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)
-\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>
-</math>.
+This appears to be more powerful, but it follows from
-</center>
+<cmath>
-This appears to be more powerful, but it follows immediately 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)
-<center>
+</cmath>
-<math>
-\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) ==
-</math>.
-</center>
+Let <math>a_1, a_2, \ldots, a_n</math> and <math>b_1, b_2, \ldots, b_n</math> be two sequences of positive real numbers with
+<cmath>
+< m \le \frac{a_i}{b_i} \le M
+</cmath>
+for <math>1 \le i \le n</math>. Then
+<cmath>
+\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,
+</cmath>
+with equality if and only if, for some ordering of the pairs <math>(a_i,b_i) \mapsto (a_{\sigma(i)},b_{\sigma(i)})</math>, some <math>0 \le j \le n</math> exists such that <math>a_{\sigma(i)}=mb_{\sigma(i)}</math> for <math>1 \le \sigma(i) \le j</math> and <math>a_{\sigma(i)}=Mb_{\sigma(i)}</math> for <math>j+1 \le \sigma(i) \le n</math>, and
+<cmath>
+m\sum_{\sigma(i)=1}^{j}b_{\sigma(i)}^2 = M\sum_{\sigma(i)=j+1}^{n}b_{\sigma(i)}^2.
+</cmath>
+If we restrict that <math>m_1 \le a_i \le M_1</math> and <math>m_2 \le b_i \le M_2</math> for all <math>i</math>, then it's clear that for <math>a_i/b_i</math> to be <math>m=m_1/M_2</math> or <math>M=M_1/m_2</math> for all <math>i</math>, then <math>a_i=m_1 \Longleftrightarrow b_i=M_2</math> and <math>a_i=M_1 \Longleftrightarrow b_i=m_2</math>, so
+<cmath>
+m\sum_{\sigma(i)=1}^{j}b_{\sigma(i)}^2 = M\sum_{\sigma(i)=j+1}^{n}b_{\sigma(i)}^2
+</cmath>
+is equivalent to
+<cmath>
+\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*}
+</cmath>
+(When this is not an integer, the maximum occurs when <math>j</math> is either the ceiling or floor of the right-hand side.) In the special case that <math>a_ib_i = k > 0</math> is constant for all <math>i</math>, we have <math>M_1=1/m_2</math> and <math>m_1=1/M_2</math>, so here <math>j</math> must be <math>n/2</math>.
+=== Proof ===
+Note that for all <math>i</math>, we have
+<cmath>
+\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)
+</cmath>
+or
+<cmath>
+(M+m)a_ib_i \ge a_i^2+(Mm)b_i^2,
+</cmath>
+with equality if and only if <math>a_i=mb_i</math> or <math>a_i=Mb_i</math>. Summing up these inequalities over <math>1 \le i \le n</math>, we obtain from AM-GM that
+<cmath>
+\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*}
+</cmath>
+and squaring gives us the desired bound. For equality to occur, we must have <math>a_i=mb_i</math> or <math>a_i=Mb_i</math> for all <math>i</math>. If, without loss of generality, <math>a_i=mb_i</math> for <math>1 \le i \le j</math> and <math>a_i=Mb_i</math> for <math>j+1 \le i \le n</math> for some <math>0 \le j \le n</math>, then for the AM-GM to reach equality we must have (assume <math>M>m</math> since <math>M=m</math> is trivial)
+<cmath>
+\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*}
+</cmath>
 == General Form ==
-Let <math>V </math> be a [[vector space]], and let <math> \langle \cdot, \cdot \rangle : V \times V \mapsto \mathbb{R} </math> be an [[inner product]].  Then for any <math> \mathbf{a,b} \in V </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>,
-<center>
+<cmath>
-<math>
+\langle \mathbf{a,b} \rangle^2 \le \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle ,
-\langle \mathbf{a,b} \rangle^2 \le \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle
+</cmath>
-</math>,
-</center>
 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>.
@@ Line 47: / Line 93: @@
 Consider the polynomial of <math> t </math>
-<center>
+<cmath>
-<math>
+\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 .
-\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>
-</math>.
-</center>
 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.
@@ Line 57: / Line 101: @@
 We consider
-<center>
+<cmath>
-<math>
+\langle \mathbf{a-b, a-b} \rangle = \langle \mathbf{a,a} \rangle + \langle \mathbf{b,b} \rangle - 2 \langle \mathbf{a,b} \rangle .
-\langle \mathbf{a-b, a-b} \rangle = \langle \mathbf{a,a} \rangle + \langle \mathbf{b,b} \rangle - 2 \langle \mathbf{a,b} \rangle
+</cmath>
-</math>.
-</center>
 Since this is always greater than or equal to zero, we have
-<center>
+<cmath>
-<math>
+\langle \mathbf{a,b} \rangle \le \frac{1}{2} \langle \mathbf{a,a} \rangle + \frac{1}{2} \langle \mathbf{b,b} \rangle .
-\langle \mathbf{a,b} \rangle \le \frac{1}{2} \langle \mathbf{a,a} \rangle + \frac{1}{2} \langle \mathbf{b,b} \rangle
+</cmath>
-</math>.
-</center>
 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
-<center>
+<cmath>
-<math>
+\langle \mathbf{a,b} \rangle \le 1 = \langle \mathbf{a,a} \rangle^{1/2} \langle \mathbf{b,b} \rangle^{1/2} ,
-\langle \mathbf{a,b} \rangle \le 1 = \langle \mathbf{a,a} \rangle^{1/2} \langle \mathbf{b,b} \rangle^{1/2}
+</cmath>
-</math>,
-</center>
 with equality when <math>\mathbf{a} </math> and <math> \mathbf{b} </math> may be scaled to each other, as desired.
+=== Proof 3 ===
+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>
 === 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 <math> f,g : [a,b] \mapsto \mathbb{R} </math>,
-<center>
+<cmath>
-<math>
+\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
-\left( \int_{a}^b f(x)g(x)dx \right)^2 \le \int_{a}^b [f(x)]^2dx \cdot \int_a^b [g(x)]^2 dx
+</cmath>
-</math>,
-</center>
 with equality when there exist constants <math> \mu, \lambda </math> not both equal to zero such that for <math> t \in [a,b] </math>,
-<center>
+<cmath>
-<math>
+\mu \int_a^t f(x)dx = \lambda \int_a^t g(x)dx .
-\mu \int_a^t f(x)dx = \lambda \int_a^t g(x)dx
+</cmath>
-</math>.
-</center>
 ==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
-<center>
+*Let <math>ABC</math> be a triangle such that
-<math>
+<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
+\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 ,
-</math>,
+</cmath>
-</center>
+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.
-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>
-<center>
-<math>
 \frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}
-</math>
+</cmath>
-</center>
 is least.
@@ Line 124: / Line 171: @@
+[[Category:Algebra]]
 [[Category:Inequality]]
 [[Category:Theorems]]

Page

Toolbox

Search