Difference between revisions of "Cauchy-Schwarz Inequality"
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==Problems== | ==Problems== |
Revision as of 21:56, 4 February 2022
In algebra, the Cauchy-Schwarz Inequality, also known as the Cauchy–Bunyakovsky–Schwarz Inequality or informally as Cauchy-Schwarz, is an inequality with many ubiquitous formulations in abstract algebra, calculus, and contest mathematics. In high-school competitions, its applications are limited to elementary and linear algebra.
Its elementary algebraic formulation is often referred to as Cauchy's Inequality and states that for any list of reals and , with equality if and only if there exists a constant such that for all , or if every number in one of the lists is zero. Along with the AM-GM Inequality, Cauchy-Schwarz forms the foundation for inequality problems in intermediate and olympiad competitions. It is particularly crucial in proof-based contests.
Its linear algebraic forulation states that for any vectors and in , with equality if and only if there exists a scalar such that , or if one of the vectors is zero. This formulation comes in handy in linear algebra problems at intermediate and olympiad problems.
The full Cauchy-Schwarz Inequality is written in terms of abstract vector spaces. Under this formulation, the elementary algebraic, linear algebraic, and calculus formulations are different cases of the general inequality.
Contents
[hide]Proof
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.
Lemmas
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.
Proof 3
Consider for some scalar . Then: (by the Trivial Inequality) . Now, let . Then, we have: .
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.