SOLUTION: 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.

add a to all three parts


divide all three parts by 2ab




add b to all three parts


divide all three parts by 2ab



Put these two together:

 and 

Ignoring the first parts of each:



We use the fact that if  then 
I'll prove that if you need it.  But I'll use it here

So that applies above and we have



And since at the top we showed 
then 

So  

Edwin

In the above I used this without proof:

If  then 

Proof:



Multiply through by x, then by y, then by z



From the 1st and 3rd inequalities (we don't need the middle one),
we get



Now we divide through by xyz





Edwin

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.