Spivak Calculus 4th ed (Ch1 problem 7)

[MidgetDwarf]

Prove that if 0<a<b, then a &lt; \sqrt{ab} &lt; \frac {a+b} {2} &lt; 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.

 

[Samy_A]

Prove that if 0<a<b, then a &lt; \sqrt{ab} &lt; \frac {a+b} {2} &lt; 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.

$a < \frac {a + b} {2}$
< $a < \frac {a + b} {2}$ < $a < \frac {b + b} {2}$ =b. Therefore a<b .

Here something went wrong.

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

 

[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.

 

[MidgetDwarf]

Here something went wrong.

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

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

 

[Samy_A]

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.

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:

< $a < \frac {a + b} {2}$ < $a < \frac {b + b} {2}$ =b. Therefore a<b

$\frac {a + b} {2}$ < $a$ is obviously wrong, probably a typo.

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

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.

 

[MidgetDwarf]

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.

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.

 

[Mark44]

Prove that if 0<a<b, then a &lt; \sqrt{ab} &lt; \frac {a+b} {2} &lt; 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.

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

 

[MidgetDwarf]

Thanks Mark for informing.