Question 1208897
<pre>

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

{{{0<a<b}}}
add a to all three parts

{{{a<2a<a+b}}}
divide all three parts by 2ab
{{{a/(2ab)<(2a)/(2ab)<(a+b)/(2ab)}}}
{{{1/(2b)<1/b<(a+b)/(2ab)}}}

{{{0<a<b}}}
add b to all three parts

{{{b<a+b<2b}}}
divide all three parts by 2ab
{{{b/(2ab)<(a+b)/(2ab)<(2b)/(2ab)}}}
{{{1/(2a)<(a+b)/(2ab)<1/a}}}

Put these two together:

{{{1/(2b)<1/b<(a+b)/(2ab)}}} and {{{1/(2a)<(a+b)/(2ab)<1/a}}}

Ignoring the first parts of each:

{{{1/b<(a+b)/(2ab)<1/a}}}

We use the fact that if {{{0<x<y<z}}} then {{{matrix(1,2,0<1/z<1/y<1/x,"")}}}
I'll prove that if you need it.  But I'll use it here

So that applies above and we have

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

And since at the top we showed {{{1/h}}}{{{""=""}}}{{{(a+b)/(2ab)}}}
then {{{h=(2ab)/(a+b)}}}

So {{{a<h<b}}} 

Edwin</pre>