If #a/b >1#, and #c# is a positive integer, prove that #a/b>(a+c)/(b+c)# ?

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Answer 1 · 2017-07-22T19:24:58

See below.

Assuming #a > 0, b > 0#, if #a/b > 1 rArr a > b# then

#ab + ac > ab + bc# or

#a(b + c) > b(a+c) rArr a/b > (a+c)/(b+c)#

Answer 2 · 2017-07-22T19:29:41

The assertion fails if #a, b < 0#

The assertion fails if #a, b < 0#

Consider #a=-3#, #b=-2# and #c=1#

Then:

#a/b = (-3)/(-2) = 3/2 > 1#

#(a+c)/(b+c) = (-3+1)/(-2+1) = (-2)/(-1) = 2 > 3/2#