Question

How to prove that a/b 0 means that a/b<(a+c)/(b+d) 10 years ago

Answer 1

Some background:
Given all numbers positive and\[ \frac{a}{b} < \frac{c}{d}\]
if we multiply both sides by b, the relation remains true. ditto for d. thus
\[ ad < bc\]

It is not obvious how to use this observation, but let's work backwards.
The claim is
\[ \frac{a}{b} < \frac{a+c}{b+d} \]
as before "cross multiply" i.e. multiply both sides by the denominators:
\[ ab+ad < ab +bc \]
subtracting ab from both sides, and we have \[ ad <bc\]
which implies \( \frac{a}{b} < \frac{c}{d}\)

so we know how to proceed:
\[ \frac{a}{b} < \frac{c}{d} \implies ad<bc \implies ab+ad< ab+bc \\
\implies a(b+d) < b(a+c) \implies \frac{a}{b} < \frac{a+c}{b+d}
\]
we can do a similar thing for the upper bound

Answer 2

Given:
a/b < c/d; a,b,c,d>0
We know that if a/b = c/d then ad = bc; =>a/b = (a+c)/(b+d). when you cross multiply you get the same result ad = bc.
This hold true for the inequality since a,b,c,d>0
That is a/b<c/d => a/b < (a+c)/(b+d) .......(1)
Now, we want to see whether (a+c)/(b+d) < c/d
Let us assume (a+c)/(b+d)>c/d.
When you cross multiply you get ad>bc, which is contradictory to our initial assumption.
Therefore, our assumption is wrong.
It implies (a+c)/(b+d)<c/d ........(2)
Combining (1) and (2), we get
a/b<(a+c)/(b+d)<c/d
Hence, proved.