<s>[INST] 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.
 [/INST] 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.
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