Question 1208896
<pre>

0 < a < b

Multiply through by a:

{{{a*0 < a*a < a*b}}}

{{{0 < a^2 < ab}}}

Nonnegative square roots are in the same order of inequality as their squares. 
So we take nonnegative square roots:

{{{0 < a < sqrt(ab)}}}

0 < a < b

Multiply through by b:

{{{a*b < b*b}}}

{{{a*b < b^2}}}

Nonnegative square roots are in the same order of inequality as their squares. 
So we take nonnegative square roots:

{{{sqrt(ab)<b}}}

{{{0 < a < sqrt(ab)}}} and {{{sqrt(ab)<b}}}

Therefore {{{matrix(1,2,0 < a<sqrt(ab)< b,"")}}}

Edwin</pre>