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

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Homework Help Overview

The discussion revolves around proving the inequality \( a < \sqrt{ab} < \frac{(a+b)}{2} < b \) under the condition that \( 0 < a < b \). The subject area includes inequalities and properties of square roots, likely within the context of calculus.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss starting points for proving the first part of the inequality, specifically \( a < \sqrt{ab} \), and some express uncertainty about how to approach proving \( \frac{(a+b)}{2} < b \). There is mention of using properties of inequalities and square roots, as well as references to established mathematical inequalities.

Discussion Status

Some participants have provided insights and suggested methods for proving parts of the inequality, while others are questioning the clarity of the approaches and the assumptions being made. There is an ongoing exploration of the relationships between the terms in the inequality.

Contextual Notes

Participants reference a specific calculus textbook, indicating a focus on rigorous proof. There are also mentions of previous posts being deleted, which may affect the continuity of the discussion.

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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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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