# Intro Analysis - Real Numbers - Inequality proof

1. Jan 6, 2010

### brntspawn

For 0<x<y, show that x<$$\sqrt{xy}$$<1/2(x+y)<y

I have no difficulty showing that x<$$\sqrt{xy}$$ and 1/2(x+y)<y. I am having difficulty with $$\sqrt{xy}$$<1/2(x+y).

x<y
xx<xy
x$$^{2}$$<xy
x<$$\sqrt{xy}$$

and

x<y
x+y<y+y
x+y<2y
1/2(x+y)<y

2. Jan 6, 2010

What do you know about the sign of $$(\sqrt y - \sqrt x)^2$$?

3. Jan 6, 2010

### brntspawn

$$(\sqrt y - \sqrt x)^2$$>0

From this I can get:
x<y
$$\sqrt{x}$$<$$\sqrt{y}$$
$$\sqrt{x}$$-$$\sqrt{x}$$<$$\sqrt{y}$$-$$\sqrt{x}$$
0<$$\sqrt{y}$$-$$\sqrt{x}$$
0<($$\sqrt{y}$$-$$\sqrt{x}$$)^{2}
0<y-2$$\sqrt{xy}$$+x
2$$\sqrt{xy}$$<y+x
$$\sqrt{xy}$$<1/2(y+x)

Assuming I am correct with this I have two more questions:
1. How did you know to work with $$(\sqrt y - \sqrt x)^2$$? Is this something I would pick up with more experience, or should I have worked backwards and manipulated it until I got back to x<y?
2. I am still relatively new to proofs, what would be the best way to put all this information together into one proof?

4. Jan 6, 2010

Your question 1: I could say experience, or work backwards. I forget (it's in the distant past) where I first saw this. It is a special case of the "arithmetic/geometric mean" inequality: I don't like using wikipedia, but a brief link is

http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means

Your problem corresponds to the case $$n = 2$$.

Question 2: Asking about the 'best way' to organize a proof can lead to an almost infinite number of suggestions, so don't be surprised if that happens in this case. I'm in the camp that says the simplest, most direct, method, is what I want to see from students, as it shows they not only understand the concepts but are able to organize the required steps in the correct order. Here, if I can be bold enough to suggest, organize your work in three stages:
First, show the left-most inequality
Second, show the inequality you just worked through
Third, show the right-most inequality

This means you've just demonstrated that:
a < b, b < c, and c < d

since inequalities are transitive, you can put these together to show a < b < c < d.

A good part of the point of an introductory analysis course (since that's the title of your post, I assume that's your class) is training in constructing solid, concise, thorough proofs (the analysis part is important, too :) ).

Good work - and good luck.

5. Jan 6, 2010

### brntspawn

Thanks for the help. :)