Question 1207708
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0 < a < b
a < b
1/a > 1/b 
1/b < 1/a
1/b < 1/h < 1/a ..... see note below
b > h > a 
a < h < b


Note:  1/h is the midpoint of 1/a and 1/b, due to the formula your teacher gave you.
Therefore, 1/h is guaranteed to be between those values. 
A way to prove this can be found through <a href="https://www.algebra.com/algebra/homework/Inequalities/Inequalities.faq.question.1207706.html">this question</a>


Don't forget to flip the inequality signs when applying reciprocals to all sides. 


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Tutor Edwin has a great approach. I'll paraphrase his method slightly.


Let's rewrite the given formula
1/h = (1/2)*(1/a + 1/b)
1/h = (1/2)*(b/(ab) + a/(ab))
1/h = (1/2)*( (a+b)/(ab) )
1/h = (a+b)/(2ab)
h = 2ab/(a+b)
I'll come back to this later.


Then,
a < b turns into a^2 < ab after multiplying both sides by a
a < b turns into ab < b^2 after multiplying both sides by b
Both a and b are positive, so there's no inequality sign flips when multiplication is applied to both sides.
a^2 < ab  &  ab < b^2 combine to a^2 < ab < b^2


and finally
a^2 < ab < b^2
a^2+ab < ab+ab < b^2+ab ...... adding ab to all sides
a^2+ab < 2ab < b^2+ab
a(a+b) < 2ab < b(a+b)
a < 2ab/(a+b) < b ........... dividing all sides by (a+b)
a < h < b


Since a > 0 and b > 0, we can be certain that a+b > 0 as well of course.
a+b > 0 means dividing both sides by (a+b) will not flip the inequality sign.
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