How to Prove AM > GM for Two Positive Numbers?

Thread starter jake_at
Start date Jan 8, 2004

AI Thread Summary

To prove that the arithmetic mean (AM) of two positive numbers exceeds the geometric mean (GM), start with two distinct positive numbers, a and b. The formula for AM is (a+b)/2, while GM is √(ab). By manipulating the equation, it can be shown that AM is greater than or equal to GM using the expression AM = (√(a) - √(b))²/2 + √(ab). This demonstrates that the difference between AM and GM is non-negative, confirming the inequality. Thus, the proof establishes that AM > GM for two positive numbers.

Jan 8, 2004

jake_at

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.

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Jan 8, 2004

himanshu121

Let us take two numbers a and b

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

clearly AM>=GM

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