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.

Algebra.Com
Question 1132161: Show that in general, if a and b are positive and then ,
where x and y can be any positive integers.
Answer by ikleyn(52814)   (Show Source): You can put this solution on YOUR website!
.
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.

Solved.



RELATED QUESTIONS

If Ax + By = Cz, where A, B, C, x, y and z are positive integers and x, y and z are all... (answered by Edwin McCravy)
Show that the quadratic function {{{y=ax^2+bx+c}}} can be written in the form... (answered by fcabanski)
The expression 6y^2-y-51 can be rewritten as (3Ay+B)(y-C), where A, B, and C are positive (answered by Boreal,ikleyn,MathTherapy)
If y= A/x + Bx where A and B are constants, show that x²dy/dx... (answered by Edwin McCravy,ikleyn)
With just the equation y=ax^2+bx+c and a parabola without any points, how do I determine... (answered by tommyt3rd)
If a and c are both positive, what can be said about the graph of... (answered by jim_thompson5910)
Suppose that A, B, and C are positive constants and that x+y= C. Show that the minimum... (answered by Edwin McCravy)
if ax²+bx+c=0, where a, b, and c are integers 
and its two solutions have to be... (answered by AnlytcPhil)

 A^x + B^y = C^z, prove that when A, B and C are positive integers, and x, y and z are... (answered by Alan3354)