SOLUTION: Show that in general, if a and b are positive and {{{a < b,}}} then {{{a/(ax + y) < b/(bx + y)}}}, where x and y can be any positive integers.

The standard way to prove such statements is to start from the given inequality and to transform it to the simplest form.

using EQUIVALENT transformations.


If the final simplified inequality is TRUE, then the starting inequality is TRUE, too, since you can reverse the chain 
of equivalent transformations.


So, we start from  

     < .                   (1)


Multiply both sides by the positive number  (ax+y)*(bx+y).  Then inequality (1) is EQUIVALENT to

    a*(bx+y) < b*(ax+y).               (2)


Inequality (2) is EQUIVALENT to

    abx + ay < abx + by                (3)


which is equivalent to

    ay < by                            (4)


which with positive "y" is equivalent to

    a < b,                             (5)


which is given.


So, the reversed chain of VALID inequalities   (5) ---> (4) ---> (3) ---> (2) ---> (1)  proves the statement.