Difference between revisions of "Cauchy-Schwarz Inequality"

m (categories)
m (cosmetics)
Line 4: Line 4:
  
 
For any real numbers <math> a_1, \ldots, a_n </math> and <math> b_1, \ldots, b_n </math>,
 
For any real numbers <math> a_1, \ldots, a_n </math> and <math> b_1, \ldots, b_n </math>,
<center>
+
<cmath>
<math>
+
\biggl( \sum_{i=1}^{n}a_ib_i \biggr)^2 \le \biggl(\sum_{i=1}^{n}a_i^2 \biggr) \biggl(\sum_{i=1}^{n}b_i^2 \biggr),
\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>,
 
</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>.
 
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>.
  
Line 20: Line 18:
  
 
Let <math> a_1, \ldots, a_n </math> and <math> b_1, \ldots, b_n </math> be [[complex numbers]].  Then
 
Let <math> a_1, \ldots, a_n </math> and <math> b_1, \ldots, b_n </math> be [[complex numbers]].  Then
<center>
+
<cmath>
<math>
+
\biggl| \sum_{i=1}^na_ib_i \biggr|^2 \le \biggl(\sum_{i=1}^{n}|a_i^2| \biggr) \biggl( \sum_{i=1}^n |b_i^2| \biggr) .
\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
+
\biggl| \sum_{i=1}^n a_ib_i \biggr| ^2 \le \biggl( \sum_{i=1}^n |a_i| \cdot |b_i| \biggr)^2 \le \biggl(\sum_{i=1}^n |a_i^2| \biggr) \biggl( \sum_{i=1}^n |b_i^2| \biggr).
<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 )
 
</math>.
 
</center>
 
  
 
== General Form ==
 
== 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>.
 
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 45: Line 37:
  
 
Consider the polynomial of <math> t </math>
 
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.
 
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 55: Line 45:
  
 
We consider
 
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
 
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
 
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.
 
with equality when <math>\mathbf{a} </math> and <math> \mathbf{b} </math> may be scaled to each other, as desired.
  
Line 77: Line 61:
  
 
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>,
 
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>,
 
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==
 
==Problems==
 +
 
===Introductory===
 
===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]])
 
*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]])
 +
 
===Intermediate===
 
===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]])
 
([[2002 USAMO Problems/Problem 2|Source]])
 +
 
===Olympiad===
 
===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
 
*<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}
 
\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}
</math>
+
</cmath>
</center>
 
 
 
 
is least.
 
is least.
  

Revision as of 13:50, 9 April 2008

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.

Elementary Form

For any real numbers $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$, \[\biggl( \sum_{i=1}^{n}a_ib_i \biggr)^2 \le \biggl(\sum_{i=1}^{n}a_i^2 \biggr) \biggl(\sum_{i=1}^{n}b_i^2 \biggr),\] with equality when there exist constants $\mu, \lambda$ not both zero such that for all $1 \le i \le n$, $\mu a_i = \lambda b_i$.

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 $\left( ||\mathbf{a}|| \cdot ||\mathbf{b}|| \cos\theta \right)^2$. The right hand side of the inequality is equal to $\left( ||\mathbf{a}|| \cdot ||\mathbf{b}|| \right)^2$. 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 \[\biggl| \sum_{i=1}^na_ib_i \biggr|^2 \le \biggl(\sum_{i=1}^{n}|a_i^2| \biggr) \biggl( \sum_{i=1}^n |b_i^2| \biggr) .\] This appears to be more powerful, but it follows from \[\biggl| \sum_{i=1}^n a_ib_i \biggr| ^2 \le \biggl( \sum_{i=1}^n |a_i| \cdot |b_i| \biggr)^2 \le \biggl(\sum_{i=1}^n |a_i^2| \biggr) \biggl( \sum_{i=1}^n |b_i^2| \biggr).\]

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)

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

Books

Invalid username
Login to AoPS