There exists $r$ and $b$ such that:
$\frac{r}{r+b} \cdot \frac{r-1}{r+b-1} = \frac{1}{2}$
I understand that from this, we can imply that:
$\frac{r}{r+b}\geq \frac{r-1}{r+b-1}$
But somehow the solution deduces without explanation that:
$\left(\frac{r}{r+b} \right)^2 > \frac{1}{2} > \left(\frac{r-1}{r+b-1} \right)^2$
I can't understand why the squared sign is introduced and how these two squared fractions are related to $\frac{1}{2}$
$xy = {1 \over 2}$
Also, you already known that $x > y$. This has nothing to do with the given equality. It's simply because if you subtract 1 from both the numerator and denominator of a fraction, you get a smaller fraction.
$x > y$ implies $x^2 > xy = {1 \over 2}$ (multiplying both sides by $x$)
$y < x$ implies $y^2 < xy = {1 \over 2}$ (multiplying both sides by $x$)
You have a the desired inequality.
I have assumed $x,y>0$ here.