Question 1207708
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I am tutor Edwin. I have two screen names (Edwin McCravy) and (mccravyedwin)
Here is how I came up with my proof above. 

Here is what I did on scratch paper, not something I would post,
or hand in if I were a student.

I simplified this, getting it down to a "flippable" fraction:

{{{1/h = (1/2)((1/a) + (1/b)^"")}}}

{{{1/h = 1/(2a) + 1/(2b)}}}

{{{1/h = (b+a)/2ab}}}

I flipped both sides:

{{{h = (2ab)/(b+a)}}}

Then (still on scratch paper), I started with the answer (which
is 'illegal', but you can keep your scratch paper a secret.  Your
teacher does not have to see it.  Anyway, I worked backwards from
the answer, even though I had no right to do so. LOL

{{{a}}}{{{""<""}}}{{{h}}}{{{""<""}}}{{{b}}}

{{{a}}}{{{""<""}}}{{{(2ab)/(b+a)}}}{{{""<""}}}{{{b}}}

Then I multiplied through by (a+b), reversed from (b+a)

{{{a(a+b)}}}{{{""<""}}}{{{2ab}}}{{{""<""}}}{{{b(a+b)}}}

{{{a^2+ab)}}}{{{""<""}}}{{{2ab}}}{{{""<""}}}{{{ab+b^2)}}}

Then I subtracted ab from all three sides:

{{{a^2)}}}{{{""<""}}}{{{ab}}}{{{""<""}}}{{{b^2)}}}

Then I noticed I could make two inequalities out of this:

{{{a^2)}}}{{{""<""}}}{{{ab}}} and {{{ab}}}{{{""<""}}}{{{b^2)}}}

I divided the first through by a, and the second one through by b,

{{{a)}}}{{{""<""}}}{{{b}}} and {{{a}}}{{{""<""}}}{{{b)}}}

They're both the same. And they're something that was given!!! 

Now I must remind you that I had no right to start with the answer
and work backwards.  However, by doing so anyway, I was able to reverse
the process and get a process that I did have a right to do. 

Now if you will compare what I just did with what I did above, you will 
see how I came up with my proof.  That is a method that sometimes helps
you to see how to come up with a proof.  On scratch paper, start with
the answer, work backwards, then see if you can reverse the steps.  Then
toss your scratch paper in the waste basket, and your teacher will never
know. [He/she might suspect though. LOL]

Edwin</pre>