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

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SUMMARY

The discussion focuses on proving the inequality \( a < \sqrt{ab} < \frac{(a+b)}{2} < b \) under the condition \( 0 < a < b \). The initial proof begins with the relationship \( a < b \) and derives \( a < \sqrt{ab} \) by squaring both sides. The second part of the proof involves demonstrating that \( \frac{(a+b)}{2} \geq \sqrt{ab} \), which is a specific case of the Arithmetic Mean-Geometric Mean Inequality. The discussion emphasizes the use of algebraic manipulation and the properties of inequalities.

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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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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?

\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}&gt;} 0

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

And the last equality should be easy. :)
 
Last edited by a moderator:
VietDao29 did you mean:

(\sqrt{x}-\sqrt{y})^2 \geq 0 ?
 
njama said:
VietDao29 did you mean:

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

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

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