Proving the Geometric Mean Inequality

[zeion]

Homework Statement

Given that 0 <or equal to a <or equal to b, show that

a <or equal to sqrt(ab) <or equal to (a+b)/2 <or equal to b

The number sqrt(ab) is called the geometric mean of a and b

Homework Equations

Not sure -_-

The Attempt at a Solution

a = sqrt(aa) <or equal to sqrt(ab) <or equal to sqrt(bb) = b
a = (a+a)/2 <or equal to (a+b)/2 <or equal to (b+b)/2 = b

Not sure what to do now -_-

 

[skeeterrr]

There are 3 things you need to show:

a < root ab
root ab < (a+b)/2
(a+b)/2 < b

So you have the given condition 0 < a < b (Let's assume that < or > also includes the equal to, since I don't know how to use the notations)

since 0 < a < b, a < b as well

try working around a < b

PS. is this from the MAT137 problem set?

 

[tnutty]

a <= b;
a*a <= b*a;
sqrt(a*a) <= sqrt(b*a)
a <= sqrt(b*a)

use similar logic for the other ones.

 

[zeion]

I'm not sure how to prove sqrt(ab) <= (a+b)/2

And yes this is for MAT137 -_-

 

[Dick]

You know both sides are positive or zero. You can square both sides and the inequality still holds. Can you see the next step?

 

[zeion]

I know that b >= a >= 0
So sqrt(b) >= sqrt(a) >= 0
I know that b - a >= 0
So sqrt(b) - sqrt(a) >= 0
I know that sqrt of any number is >= 0
So sqrt(sqrt(b) - sqrt(a)) >= 0
Now I square now sides twice
(sqrt(b) - sqrt(a))^2 >= 0
Expand
b - 2sqrt(ab) + a >= 0
a + b / 2 >= sqrt(ab)

I'm pretty sure this is right.. though I only came this part with someone else' help.
Now I'm wondering if there is any other way to solve this and how I can learn to think this way and be able to solve other similar questions.. any advice would be appreciated, thanks.