# Homework Help: Need help with proving something?

1. Sep 13, 2009

### monolithic

1. The problem statement, all variables and given/known data

0 < a < b

Must prove that a < sqrt(ab) < (a+b)/2 < b

2. Relevant equations

Well, relevant info. I am using Spivak's calculus book, where we have to prove everything.

3. The attempt at a solution

I'm not sure if I'm doing this right but here's my start.

I'll start with a < b. Multiply both sides by a, and you get a times a < a times b.

Then you get a^2 < ab

Then square root of a^2 < square root of ab.

a < square root of ab.

So I think i have the first part of that, but I'm not sure where to start for a+b/2?

2. Sep 13, 2009

### VietDao29

$$\frac{a + b}{2} \geq \sqrt{ab}$$

This is actually a special case of http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#The_inequality" (the case for n = 2, 2 terms). This case (n = 2) can be easily proven by isolating everything to one side, and use the fact that:

$$(x - y) ^ 2 = x ^ 2 - 2xy + y ^ 2 \geq 0$$

And for $$x \neq y$$, we have the equality:

$$(x - y) ^ 2 = x ^ 2 - 2xy + y ^ 2 {\color{red}>} 0$$

Let's see if you can take it from here.

And the last equality should be easy. :)

Last edited by a moderator: Apr 24, 2017
3. Sep 13, 2009

### njama

VietDao29 did you mean:

$$(\sqrt{x}-\sqrt{y})^2 \geq 0$$ ?

4. Sep 13, 2009

### VietDao29

No, I mean x, y; x, y in general.

5. Sep 13, 2009

### elduderino

My post got deleted?!! Im new to the forum here.. someone please shed light.

6. Sep 14, 2009

### njama

If he use x=a, and y=b in the inequality that I posted, it will be very easy to prove it.

7. Sep 14, 2009

### HallsofIvy

You gave a complete solution rather than helping hints.