Homework Help: AM-GM Inequality - Troubles with an example

1. Jan 21, 2013

Kolmin

1. The problem statement, all variables and given/known data

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

2. Relevant 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?

3. 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.

2. Jan 21, 2013

Integral

Staff Emeritus
Try moving the parens around on the RHS, you should be able to see the equality.

3. Jan 21, 2013

Kolmin

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

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.

4. Jan 21, 2013

Kolmin

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?

5. Jan 25, 2013

Andrax

Random question here : what ensure that the solution 3 is the LEAST possible solution?