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For 0 < a < b, let h be defined by
1/h = (1/2)[(1/a) + (1/b)].
Show that a < h < b.
Note: The number h is called the harmonic mean of a and b.
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Let' simplify
= = .
Since = , it implies that h = .
Now we want to prove that
a < < b. (*)
So, your starting inequality is
a < b. (1)
In (1), multiply both sides by positive number "a".
You will get an equivalent inequality
a^2 < ab. (2)
Next step, in (1), multiply both sides by positive number "b".
You will get an equivalent inequality
ab < b^2. (3)
From (2) and (3), you have this compound inequality
a^2 < ab < b^2. (4)
Add ab to all three terms in inequality (4). You will get an equivalent inequality
a^2 + ab < ab + ab < b^2 + ab. (5)
Rewrite it equivalently this way
a*(a+b) < 2ab < b*(a+b). (6)
In (6), divide all three sides by positive real number a+b. You will get an equivalent inequality
a < < b. (7)
Compare it with (*) : inequality (7) is what you want to prove.
Solved.