How to Prove AM > GM for Two Positive Numbers?

  • Context: Undergrad 
  • Thread starter Thread starter jake_at
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on proving that the arithmetic mean (AM) of two distinct positive numbers exceeds the geometric mean (GM) of the same numbers. The formula for AM is given as AM = (a + b) / 2, while GM is represented as GM = √(ab). The proof utilizes the inequality AM ≥ GM, which is established through the manipulation of the expression (√a - √b)², demonstrating that AM is always greater than or equal to GM for any two positive numbers a and b.

PREREQUISITES
  • Understanding of arithmetic and geometric means
  • Basic algebraic manipulation skills
  • Familiarity with inequalities in mathematics
  • Knowledge of positive real numbers
NEXT STEPS
  • Study the proof of the AM-GM inequality in detail
  • Explore applications of AM-GM in optimization problems
  • Learn about other inequalities in mathematics, such as Cauchy-Schwarz
  • Investigate the implications of AM-GM in calculus and analysis
USEFUL FOR

Students of mathematics, educators teaching inequalities, and anyone interested in mathematical proofs and their applications in various fields.

jake_at
Messages
4
Reaction score
0
i'm having some trouble with this question on GP/APs. can anybody show me how to do it, and explain? thanks.

:) J.

The Culprit:
Prove that the arithmetic mean of two different positive numbers exceeds the geometric mean of the same two numbers.
 
Mathematics news on Phys.org
Let us take two numbers a and b

AM= (a+b)/2 = [tex]\frac{(\sqrt{a}-\sqrt{b})^2}{2} + \sqrt{ab}[/tex]

clearly AM>=GM
 

Similar threads

  • · Replies 35 ·
2
Replies
35
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K