How does squaring these multiplicands permit inequality comparison to their multiple? This question pertains to Mosteller's classic book Fifty Challenging Problems in Probability. Specifically, this in regards to an algebraic operation Mosteller performs in the solution to the first question, entitled "The Sock Drawer."
Mosteller writes:

Then we require the probability that both are red to be $\frac{1}{2}$, or $$\frac{r}{r+b}*\frac{r-1}{r+b-1}=\frac{1}{2}\text{.}$$
  …
Notice that $$\frac{r}{r+b}\gt\frac{r-1}{r+b-1}\text{, for $b > 0$.}$$ Therefore we can create the inequalities $$\left(\frac{r}{r+b}\right)^2 \gt \frac 12 \gt \left(\frac{r-1}{r+b-1}\right)^2$$

Despite much staring, and not knowing what to Google, I am stumped! In that last step, how does he do that‽
Many thanks,
James
 A: You have two positive numbers, which I'll call $x$ and $y$, with
$xy=\frac12$ and $x>y$.
Then
$$x^2>xy>y^2$$
which is the same as
$$x^2>\frac12>y^2.$$
A: \begin{eqnarray*}
\frac{r}{r+b} \color{blue}{\frac{r}{r+b}} > \frac{r}{r+b} \color{blue}{\frac{r-1}{r+b-1}}=\frac{1}{2} =\color{orange}{\frac{r}{r+b}} \frac{r-1}{r+b-1}>\color{orange}{\frac{r-1}{r+b-1}} \frac{r-1}{r+b-1}.
\end{eqnarray*}
A: If $x>y$, then $x*x > xy$. But $xy = \frac12$ so $x^2 > \frac12$.
Similar for $\frac12 > y^2$.
A: In a product of two (positive) things, if I replace one of them with something bigger, I get something bigger and if I replace one of them with some thing smaller I get something smaller.  So, if in 
$$\frac{r}{r+b} \cdot \frac{r-1}{r+b-1}=\frac{1}{2}$$
I replace $\frac{r}{r+b}$ with something smaller, say $\frac{r-1}{r+b-1}$, I get 
$$\frac{r-1}{r+b-1} \cdot \frac{r-1}{r+b-1} < \frac{r}{r+b} \cdot \frac{r-1}{r+b-1}=\frac{1}{2}  \text{,}$$ or
$$\left( \frac{r-1}{r+b-1} \right)^2 < \frac{1}{2}  \text{.}$$
Substituting the larger of the two for the second term gives the other inequality.
