AM-GM Inequality - Troubles with an example

AI Thread Summary
The discussion revolves around the application of the AM-GM inequality to find the least possible value of the expression a + 1/(b(a-b)) for real numbers a and b, where a > b > 0. Participants express confusion about the manipulation of the equation and the rationale behind using three specific elements in the inequality. Clarifications are sought regarding the necessity of the chosen elements and how they relate to the equality in the AM-GM context. The conversation highlights a common struggle with understanding inequalities and the conditions for achieving the least value. Ultimately, the importance of recognizing the structure of the inequality and its components is emphasized.
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Homework Statement



Let ##a## and ##b## real numbers such that ##a>b>0##.

Determine the least possible value of ##a+ \frac{1}{b(a-b)}##

I took this example from page 3 of this paper

Homework Equations



In the article previously linked, explaining the example, the author writes down:

a+ \frac{1}{b(a-b)}=(a-b)+b+\frac{1}{b(a-b)}

Now, where does that come from?

The Attempt at a Solution



As the title says, at least I am aware of what the topic is... (!). So everything moves around the AM-GM inequality.

\frac{a_1 + \dots + a_n}{n} \geq \sqrt[n]{a_1 \dots a_n}

I have to admit I have some troubles figuring out what's going on here, so it's not a matter of solving something, it's more about showing why I don't see the solution.

I tried to manipulate a bit the first formula, but it's not about that I guess, cause I really cannot see how the equality in 2. stands.

So, I am looking forward to any feedback. Thanks a lot.
 
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Try moving the parens around on the RHS, you should be able to see the equality.
 
Integral said:
Try moving the parens around on the RHS, you should be able to see the equality.

Jeez. That's really bad... completely mathematically blind.
I am quite ashamed of myself. :redface:

Now that I see that 2+2=4, I have some problems that I am afraid will be challenging like 3+3=?.

There are some questions I have related with other things I don't understand.

1) Why do we need to use that trick I didn't understand? Why do we need it?
2) Why in the LHS we have a 3 before the root? From that 3, it seems that the GM formula should be ##n \sqrt[n]{a_1 \dots a_n}##.
3) In other words, why do we need 3 elements?
4) Last dumb question, why are ##b##, ##(a-b)## and ##1/b(a-b)## our three elements? Shouldn't they be ##a## and ##b##?

I guess that now it's clear that I have a real problem with inequalities.
 
Ok, correct me if I am wrong.
I think I see now what's going on here.

1) we build up that trick I couldn't see to get rid of everything under the root in the GM side;
2) we have a 3 in the LHS cause it comes from the AM in the RHS;
3) yeah, we do need three elements to implement that trick on 1.;
4) no, the elements are indeed three.

Did I guess it right? :smile:
 
Random question here : what ensure that the solution 3 is the LEAST possible solution?
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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