= Solution
Let $C=\overline B_1\cap B_2$. Under replacement its joint <probability> is
$$
P(R\cap C)=\frac{N-N_B}{N}\,r\,\frac{N_B}{N},
$$
whereas without replacement it is
$$
P(\overline R\cap C)=\frac{N-N_B}{N}\,(1-r)\,\frac{N_B}{N-1}.
$$
In the latter case the first removed ball is non-blue, so all $N_B$ blue balls remain. Applying <Bayes' theorem> and cancelling the positive common factors yields the <posterior replacement probability after different-colour draws>:
$$
\boxed{g(r)=P(R\mid C)=\frac{r/N}{r/N+(1-r)/(N-1)}=\frac{r(N-1)}{N-r}.}
$$
It does not depend on $N_B$. To prove the requested bound directly,
$$
\boxed{r-g(r)=\frac{r(1-r)}{N-r}\ge0,\qquad0\le r\le1.}
$$
Equality holds at $r=0$ and $r=1$, and the inequality is strict between them. Its sketch joins $(0,0)$ to $(1,1)$, stays below the diagonal, and is increasing and convex because
$$
g'(r)=\frac{N(N-1)}{(N-r)^2}>0,\qquad g''(r)=\frac{2N(N-1)}{(N-r)^3}>0.
$$
The interpretation is that removing a non-blue ball increases the chance of blue on the second draw; seeing the two different colours therefore favours the no-replacement explanation.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-1-2019-q16-replacement-posterior.png]
{title=Posterior replacement probability lies below the original replacement probability}
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