AM-GM Inequality: Prove 0 <= A <= B

[SolidSnake]

Homework Statement

Show that:
0 <= A <= B

$$A \leq \sqrt{AB} \leq (A + B)/ 2 \leq B$$

Homework Equations

root AB = geometric mean
(A + B)/ 2 = arithmetic mean

<= means less then or equal to.

The Attempt at a Solution

I managed to come up with something for the (A + B)/ 2 part.

a < = (a+a)/2 <= (a+b)/2 <= (b+b)/2 <= b

that shows that (a+b)/2 is less then b but greater then a. i can't figure out how to show that with root ab.

Any help would be great!

 

[slider142]

Hint: What do you know about (B - A)^2 ?

 

[SolidSnake]

i added something i forgot to mention earlier.

(B-A)^2 = B^2 -2ab + A^2
Assuming that (B-A)^2 = 0
And if i were to rearrange it i'd get B^2 +A^2 = 2ab
And then if divided the 2 and took the root i'd be left with B+A/root 2 = root AB

That root 2 is still throwing me off tho. :S

 

[slider142]

i added something i forgot to mention earlier.

(B-A)^2 = B^2 -2ab + A^2
Assuming that (B-A)^2 = 0
And if i were to rearrange it i'd get B^2 +A^2 = 2ab
And then if divided the 2 and took the root i'd be left with B+A/root 2 = root AB

That root 2 is still throwing me off tho. :S

Why the equality? Since C^2 >= 0 for all C, we know that (B - A)^2 >= 0.
So as you arranged it, we have B^2 + A^2 >= 2AB
That is equivalent to A^2 + 2AB + B^2 >= 2AB + 2AB = 4AB.
Can you see where to go from here?

 

[SolidSnake]

:O the light bulb just went on. Thanks a lot :).