If $$\frac {a}{a+b}<\frac{a'}{a'+b'}$$ then how can I show that $$\frac {a}{a+2b}<\frac{a'}{a'+2b'}\ \forall\ a,b,c>0$$
I tried puitting in a constant k so $$\frac {a}{a+b}=k*\frac{a}{a+2b}$$ and I get $$k=\frac{a+b}{a+2b}$$ and I dont know where from there.
On
Looks like it isn't true even for all numbers positive.
$$ \frac{30}{6}< \frac{6}{1} $$
but
$$ \frac{130}{106}> \frac{106}{101} $$
On
I can show that the relation is not general. Define a function $f(x)=(a+x)(b'+x)-(a'+x)(b+x)=ab'-a'b+(a+b'-a'-b)x$.
By the assumption (all numbers are positive), $f(0)=ab'-a'b>0$. But, under the assumption, we cannot determine the slope of $f(x)$:
slope $=a+b'-a'-b$.
Therefore, this relation is not a general one.
I'll assume everything is positive.
If $\frac {a}{a+b}<\frac{a'}{a'+b'}$, then $a(a'+b') < a'(a+b)$.
We want to know when $\frac {a}{a+2b}<\frac{a'}{a'+2b'}$ or $a(a'+2b') < a'(a+2b)$.
Let $u = a'(a+b)-a(a'+b') > 0$ and $v = a'(a+2b)-a(a'+2b')$. We want to when $v > 0$.
Since $u = a'b-ab' > 0$, $a'b > ab'$.
We have $v = (a'(a+b)+a'b)-(a(a'+b')+ab') = a'(a+b)-a(a'+b')+a'b-ab' = u+a'b-ab' $.
Since, as shown above, $a'b > ab'$ and $u > 0$, then $v > 0$ as we wanted to show.