Please give an algebraic proof of the following inequality: $\frac{a}{a+b} \gt \frac{a-1}{a+b-1}$ for $a,b \gt 0$. For $ a,b > 0 $ and $\left(\frac{a}{a+b}\right)\cdot\left(\frac{a-1}{a+b-1}\right)=\frac{1}{3}$, show that $$\frac{a}{a+b} > \frac{a-1}{a+b-1}$$
I am not getting how to do this. Please enlighten me. What i need is an algebraic proof.
My thought, let $f(x) = \frac{x}{x+b}$, and $f'(x) = \frac{b}{(x+b)^2}$, this implies $f(x)$ is increasing, i.e, $f(x) > f(x-1)$. But can we solve it simply, I tried solving just using inequalities I can't find how.
 A: You have
$$
\frac{a}{a+b}-\frac{a-1}{a+b-1}=\frac{b}{(a+b)(a+b-1)}
$$
Given that $a,b>0$, your statement is equivalent to $a+b-1>0$.
If $a$ and $b$ are integer, as you seem to hint in the comments, then the statement is completely obvious.
For general real $a,b$, we'd like to see whether this follows from
$$
\frac{a}{a+b}\frac{a-1}{a+b-1}=\frac{1}{3}
$$
It's simpler if we set $a+b=c$, so the equality becomes
$$
3a^2-3a=c^2-c
$$
We'd like to see whether the equation $c^2-c-3a^2+3a=0$ has or not solutions satisfying $a<c<1$, which would prove the conjecture false.
The roots are
$$
\frac{1\pm\sqrt{12a^2-12a+1}}{2}
$$
We have
$$
a<\frac{1-\sqrt{12a^2-12a+1}}{2}
$$
if and only if $1-2a>\sqrt{12a^2-12a+1}$, which reduces to $0<a<(3-\sqrt{6})/6$. In this case, the root is obviously less than $1$.
Suppose $a=0.01$. Then the chosen root is $\approx0.03$ and your conjecture is false.
A: Multiplying both sides by  $\frac{a}{a+b}$ you get because of your hypothesis :
$$\frac{a}{a+b} > \frac{a-1}{a+b-1} \implies (\frac{a}{a+b})^2 > \frac{1}{3} \implies a(\frac{\sqrt{3}-1}{\sqrt{3}}) >  b\frac{1}{\sqrt{3}} \implies a>\frac{b}{\sqrt{3}-1}$$
But this is not always the case.
