Difference between revisions of "Cauchy-Schwarz Inequality"
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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>, | ||
− | < | + | <cmath> |
− | + | \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), | |
− | \ | + | </cmath> |
− | |||
− | </ | ||
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>. | ||
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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 | ||
− | < | + | <cmath> |
− | + | \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) . | |
− | \ | + | </cmath> |
− | + | This appears to be more powerful, but it follows from | |
− | </ | + | <cmath> |
− | This appears to be more powerful, but it follows | + | \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). |
− | < | + | </cmath> |
− | |||
− | \ | ||
− | |||
− | </ | ||
== General Form == | == General Form == | ||
− | Let <math>V </math> be a [[vector space]], and let <math> \langle \cdot, \cdot \rangle : V \times V \ | + | 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 , | |
− | \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>. | 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>. | ||
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Consider the polynomial of <math> t </math> | 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 . | |
− | \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. | 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. | ||
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We consider | We consider | ||
− | < | + | <cmath> |
− | + | \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> |
− | |||
− | </ | ||
Since this is always greater than or equal to zero, we have | 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 . | |
− | \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 | 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} , | |
− | \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. | with equality when <math>\mathbf{a} </math> and <math> \mathbf{b} </math> may be scaled to each other, as desired. | ||
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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>, | ||
− | < | + | <cmath> |
− | + | \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 | |
− | \ | + | </cmath> |
− | </ | ||
− | |||
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>, | ||
− | < | + | <cmath> |
− | + | \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> |
− | |||
− | </ | ||
==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 | + | |
− | < | + | *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 | + | \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. |
− | 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> | |
− | < | ||
− | |||
\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF} | \frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF} | ||
− | </ | + | </cmath> |
− | |||
− | |||
is least. | is least. | ||
Revision as of 14:50, 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.
Contents
Elementary Form
For any real numbers and
,
with equality when there exist constants
not both zero such that for all
,
.
Discussion
Consider the vectors and
. If
is the angle formed by
and
, then the left-hand side of the inequality is equal to the square of the dot product of
and
, or
. The right hand side of the inequality is equal to
. The inequality then follows from
, with equality when one of
is a multiple of the other, as desired.
Complex Form
The inequality sometimes appears in the following form.
Let and
be complex numbers. Then
This appears to be more powerful, but it follows from
General Form
Let be a vector space, and let
be an inner product. Then for any
,
with equality if and only if there exist constants
not both zero such that
.
Proof 1
Consider the polynomial of
This must always be greater than or equal to zero, so it must have a non-positive discriminant, i.e.,
must be less than or equal to
, with equality when
or when there exists some scalar
such that
, as desired.
Proof 2
We consider
Since this is always greater than or equal to zero, we have
Now, if either
or
is equal to
, then
. Otherwise, we may normalize so that
, and we have
with equality when
and
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 ,
with equality when there exist constants
not both equal to zero such that for
,
Problems
Introductory
- Consider the function
, where
is a positive integer. Show that
. (Source)
Intermediate
- Let
be a triangle such that
where
and
denote its semiperimeter and inradius, respectively. Prove that triangle
is similar to a triangle
whose side lengths are all positive integers with no common divisor and determine those integers.
(Source)
Olympiad
is a point inside a given triangle
.
are the feet of the perpendiculars from
to the lines
, respectively. Find all
for which
is least.
(Source)
Other Resources
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.