Difference between revisions of "Cauchy-Schwarz Inequality"
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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. | 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 == | == Elementary Form == |
Revision as of 02:12, 19 August 2018
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.
Contents
[hide]Elementary Form
For any real numbers and , with equality when there exists a nonzero constant 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
Upper Bound on (Σa)(Σb)
Let and be two sequences of positive real numbers with for . Then with equality if and only if, for some ordering of the pairs , some exists such that for and for , and If we restrict that and for all , then it's clear that for to be or for all , then and , so is equivalent to (When this is not an integer, the maximum occurs when is either the ceiling or floor of the right-hand side.) In the special case that is constant for all , we have and , so here must be .
Proof
Note that for all , we have or with equality if and only if or . Summing up these inequalities over , we obtain from AM-GM that and squaring gives us the desired bound. For equality to occur, we must have or for all . If, without loss of generality, for and for for some , then for the AM-GM to reach equality we must have (assume since is trivial)
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)
- (APMO 1991 #3) Let , , , , , , , be positive real numbers such that . Show that
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.