Question

State arithmetic-geometric inequality theorem. Explain how it is used in deriving dual problem for a given unconstrained geometric problem.

Answer 1

there is the answer of the question

Answer 2

The Arithmetic Mean-Geometric Mean Inequality (AM-GM or AMGM) is an elementary inequality, and is generally one of the first ones taught in inequality courses.

Contents

1 Theorem

1.1 Proof

1.2 Weighted Form

2 Extensions

3 Problems

3.1 Introductory

3.2 Intermediate

3.3 Olympiad

4 See Also

Theorem

AM-GM states that for any set of nonnegative real numbers, the arithmetic mean of the set is greater than or equal to the geometric mean of the set. Algebraically, this is expressed as follows.

For a set of nonnegative real numbers , the following always holds:Using the shorthand notation for summations and products:For example, for the set , the arithmetic mean, 25, is greater than the geometric mean, 18; AM-GM guarantees this is always the case.

The equality condition of this inequalitystates that the arithmetic mean and geometric mean are equal if and only ifall members of the set are equal.

AM-GM can be used fairly frequently to solve Olympiad-level inequality problems, such as those on the USAMOand IMO.

Proof

See here: Proofs of AM-GM.

Weighted Form

The weighted form of AM-GM is given by using weighted averages. For example, the weighted arithmetic mean of  and  with  is  and the geometric is .

AM-GM applies to weighted averages. Specifically, the weighted AM-GM Inequality states that if  are nonnegative real numbers, and  are nonnegative real numbers (the "weights") which sum to 1, thenor, in more compact notation,Equality holds if and only if  for all integers  such that  and . We obtain the unweighted form of AM-GM by setting .

Extensions