How Do You Prove a < sqrt(ab) < (a+b)/2 < b?

  • Thread starter monolithic
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Please give hints instead.In summary, the problem is to prove the inequality a < sqrt(ab) < (a+b)/2 < b using Spivak's calculus book. To start, you can 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 > 0. Can you take it from here to prove the last two parts of the inequality?
  • #1
monolithic
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Homework Statement



0 < a < b

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

Homework Equations



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

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?
 
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  • #2
monolithic said:

Homework Statement



0 < a < b

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

Homework Equations



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

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?

[tex]\frac{a + b}{2} \geq \sqrt{ab}[/tex]

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:

[tex](x - y) ^ 2 = x ^ 2 - 2xy + y ^ 2 \geq 0[/tex]

And for [tex]x \neq y[/tex], we have the equality:

[tex](x - y) ^ 2 = x ^ 2 - 2xy + y ^ 2 {\color{red}>} 0[/tex]

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

And the last equality should be easy. :)
 
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  • #3
VietDao29 did you mean:

[tex](\sqrt{x}-\sqrt{y})^2 \geq 0 [/tex] ?
 
  • #4
njama said:
VietDao29 did you mean:

[tex](\sqrt{x}-\sqrt{y})^2 \geq 0 [/tex] ?

No, I mean x, y; x, y in general.
 
  • #5
My post got deleted?! I am new to the forum here.. someone please shed light.
 
  • #6
VietDao29 said:
No, I mean x, y; x, y in general.

If he use x=a, and y=b in the inequality that I posted, it will be very easy to prove it. :smile:
 
  • #7
elduderino said:
My post got deleted?! I am new to the forum here.. someone please shed light.
You gave a complete solution rather than helping hints.
 

FAQ: How Do You Prove a < sqrt(ab) < (a+b)/2 < b?

How can I prove my hypothesis?

In order to prove your hypothesis, you will need to conduct a scientific experiment. This involves creating a testable prediction based on your hypothesis, designing and conducting the experiment, and analyzing the data collected. Through this process, you can determine if your hypothesis is supported or not.

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What role does statistical analysis play in proving something?

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Communicating your findings is an important part of the scientific process. You can do this through publishing your results in a scientific journal or presenting them at a conference. It is important to clearly and accurately report your methods, results, and conclusions in order to provide evidence for your claim.

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