Question 1208897
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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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<pre>
Let' simplify

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


Since  {{{1/h}}} = {{{(1/2)*((1/a) + (1/b))}}},  it implies that  h = {{{(2ab)/(a+b)}}}.


Now we want to prove that  

    a < {{{(2ab)/(a+b)}}} < 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 < {{{(2ab)/(a+b)}}} < b.    (7)


Compare it with (*) :  inequality (7) is what you want to prove.
</pre>

Solved.