How to prove that $(\sqrt{3} + 1)b/2 < a < 1 + (\sqrt{3} + 1)b/2$? I'm starting with the assumption that $a, b > 0$, and $$\frac{a - 1}{a + b - 1} \cdot \frac{a}{a + b} = \frac{1}{3}$$
I am trying to show that $$(\sqrt{3} + 1)b/2 < a < 1 + (\sqrt{3} + 1)b/2.$$
Since $$\frac{a - 1}{a + b - 1} < \frac{a}{a + b}$$ I can write that $$\left(\frac{a - 1}{a + b - 1}\right)^2 < \frac{1}{3} < \left(\frac{a}{a + b}\right)^2$$
so $$\frac{a - 1}{a + b - 1} < \frac{1}{\sqrt{3}} < \frac{a}{a + b}$$
If I write $$\frac{a - 1}{a + b - 1} < \frac{1}{\sqrt{3}} < \frac{a}{a + b} < \frac{a}{a + b - 1}$$
I can derive the worse bound $(\sqrt{3} + 1)(b - 1)/2 < a < (\sqrt{3} + 1)(b + 2)/2$. If I could have a hint on how I can derive the one stated in the question, that would be much appreciated. And, if possible, what are some general strategies I should consider when trying to solve these kinds of problems?
Thank you!
Edit: I got this from Probability, Random Variables, and Stochastic Processes by Papoulis. The problem is the following -

If we assume that $a$ and $b$ must be integers greater than or equal to 1 does this resolve the problem with the bound?

The condition gives:
$$2a^2-2a(b+1)-b^2+b=0,$$ which gives
$$a=\frac{b+1+\sqrt{3b^2+1}}{2}$$ or
$$a=\frac{b+1-\sqrt{3b^2+1}}{2}.$$
In the first case we need to prove that
$$\frac{(1+\sqrt3)b}{2}<\frac{b+1+\sqrt{3b^2+1}}{2}<1+\frac{(1+\sqrt3)b}{2}$$ or
$$\sqrt3b<1+\sqrt{1+3b^2}<2+\sqrt3b,$$ which is obvious.
In the second case we need $$\frac{b+1-\sqrt{3b^2+1}}{2}>0,$$ which gives $$0<b<1.$$
Can you end it now?
I got that there is a problem with $$\frac{(1+\sqrt3)b}{2}<\frac{b+1-\sqrt{3b^2+1}}{2},$$ which is wrong for $b\rightarrow1^-.$
Because in this case we need to prove that:
$$\frac{1+\sqrt3}{2}\leq\frac{2-2}{2},$$ which is wrong.
If $a\geq1$ and $b\geq1$ so the second case is impossible because the second root is negative and we are done! 
