# Homework Help: Spivak Calculus 4th ed (Ch1 problem 7)

1. Feb 8, 2016

### MidgetDwarf

Prove that if 0<a<b, then $a < \sqrt{ab} < \frac {a+b} {2} < b$

Please excuse if format is messy, this is my first time writing in Latex.

Suppose 0<a<b then 0<a+O =a (1). Since a<b we can can rewrite this using (1) which is a+0<b (2).
Adding a to both sides of (2) (closed under addition) we get 2a< a+b (3). Using the multiplicative inverse of 2, (3) now becomes $a < \frac {a + b} {2}$ (4). Using the fact that a<b, $a < \frac {a + b} {2}$
< $a < \frac {a + b} {2}$ < $a < \frac {b + b} {2}$ =b. Therefore a<b .

Here is were I am confused. I am missing the step were $a < \sqrt {ab}$. Can I say that a*b<a*a=a^2 ?
Then take the square root? If this is true how would I connect it with $a < \frac {a + b} {2}$ ?

Strong criticism welcomed.

2. Feb 8, 2016

### Samy_A

Here something went wrong.

For $\sqrt {ab} \lt \frac {a+b}{2}$, you could use the following:
$(a+b)²>0,\ (a-b)²>0$.

3. Feb 8, 2016

### MidgetDwarf

I am thinking that 0<a<b

a<b
$a * a < a * b$
$\sqrt{a*a} < \sqrt{a*b}$
$a + b < \sqrt{a*b} + b < \sqrt{b*b} +b= 2*b$
(a+b)/2 < b ?

I am thinking this is correct.

4. Feb 8, 2016

### MidgetDwarf

this come from the fact that if a<b then b^2-a^2= (b-a)(b+a)?

5. Feb 8, 2016

### Samy_A

This is needlessly complicated:
$\frac{a+b}{2} \lt \frac{b+b}{2}=b$ is all you need for the last inequality.

What I meant with "something went wrong" was about the following:
$\frac {a + b} {2}$ < $a$ is obviously wrong, probably a typo.
I don't see how that helps you in proving the missing piece: $\sqrt {ab}< \frac {a + b} {2}$
You could square both sides and calculate the difference. That's where $(a-b)²>0$ will help.

6. Feb 8, 2016

### MidgetDwarf

Ahh very clever would have never thought about (a-b)^2 >0 part. By using this fact, it we just play with addition and subtraction of these terms. Thanks alot.

7. Feb 8, 2016

### Staff: Mentor

@MidgetDwarf, in future posts, please don't delete the homework template. Its use is required in problems posted in the Homework sections.

8. Feb 8, 2016

### MidgetDwarf

Thanks Mark for informing.