Proving a/b < (a+c)/(b+d) for Positive Numbers: A First Course in Calculus

[dmcharg]

Hi
I am working my way through Serge Langs A first course in Calculus and have encountered this question/proof which i am not sure how to do. Any assistance much appreciated.

Let a,b,c,d > 0 such that a/b > c/d Prove that

a/b < (a+c)/(b+d)

?

Thanks
David.

 

[matt grime]

IN which section is this? The surrounding material may well be helpful - is it near stuff on cauchy schwartz, or the triangle inequality? Or something else entirely? My experience with this type of question is that it is 'easy' with the right method, and impossible if you don't know/guess it. Any similar questions in the text near this one may well give you plenty of insight.

 

[HallsofIvy]

My first thought would be to multiply both sides of the inequality by b and b+ d.

 

[Rogerio]

Let a,b,c,d > 0 such that a/b > c/d Prove that

a/b < (a+c)/(b+d)

It is wrong.
If a/b > c/d , then
ad > bc , or
ab+ad > ab+bc , or
a(b+d) > b(a+c) , or
a/b > (a+c)/(b+d)

That's all.

 

[dmcharg]

Thanks. Yes i got the direction of the inequality the wrong way round but i see your approach. First multiply both side by b, then d and then add ab to both sides, all of which can only be done on the assumption that all values are > 0 and hence preserving the direction of the inequality.