How to Prove AM > GM for Two Positive Numbers?

[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.

 

[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

 

[tacman]

First, let's define the arithmetic mean and geometric mean of two numbers. The arithmetic mean (AM) is the sum of two numbers divided by 2, while the geometric mean (GM) is the square root of the product of the two numbers. In mathematical notation, this can be written as:

AM = (a + b)/2
GM = √(ab)

To prove that AM > GM for two positive numbers, we can use the following steps:

Step 1: Set up the inequality.
We want to prove that AM > GM, so we can start by setting up the inequality:

(a + b)/2 > √(ab)

Step 2: Square both sides.
To make the inequality easier to work with, we can square both sides of the inequality. This will not change the direction of the inequality, but it will get rid of the square root on the right side:

((a + b)/2)^2 > (√(ab))^2

Simplifying, we get:

(a + b)^2/4 > ab

Step 3: Expand the left side.
Next, we can expand the left side of the inequality using the FOIL method:

(a + b)^2 = a^2 + 2ab + b^2

Substituting this into the inequality, we get:

(a^2 + 2ab + b^2)/4 > ab

Step 4: Simplify the left side.
We can simplify the left side by dividing each term by 4:

a^2/4 + 2ab/4 + b^2/4 > ab

Simplifying further, we get:

a^2/4 + ab/2 + b^2/4 > ab

Step 5: Rearrange the terms.
To make it easier to see how AM and GM are related, we can rearrange the terms on the left side of the inequality:

a^2/4 + ab/2 + b^2/4 = (a^2 + 2ab + b^2)/4 = (a + b)^2/4

Substituting this back into the inequality, we get:

(a + b)^2/4 > ab

Step 6: Simplify the right side.
We can simplify the right side by multiplying ab by 4/4, which is essentially multiplying by 1:

(a + b)^2/4 > ab