AM-GM Inequality - Troubles with an example

[Kolmin]

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.

 

[Integral]

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

 

[Kolmin]

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.

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.

 

[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?

 

[Andrax]

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