The exercise is to determine values for $j$ which give $b(n,p,j)$ its maximum value. It begins saying, Show that:
$$b(n,p,j)=\frac{p}{q} (\frac{n-j+1}{j})b(n,p,j-1) \hspace{1cm} for \hspace{3mm} j \ge 1$$
we define $b(n,p,j)$ as:
$${{n}\choose{j}}p^jq^{n-j}$$
and $$q=1-p$$
I know how to show the first part but then it says: "Use this fact to determine the value or values of $j$ which give $b(n,p,j)$ its maximum value. The answer in the back of my book shows the first part and then gives:
$$\frac{(n-j+1)}{j}\frac{p}{q} \ge 1 \iff j\le p(n+1)$$
So $j=[p(n+1)]$ gives $b(n,p,j)$ its largest value and if it is an integer there will be two possible values of $j$: $j=p(n+1)$ and $j=p(n+1)-1$
I am having a hard time figuring out how they came up with the inequality relationship they provided. I also do not see why there are two values for $j$ and not just one.
To find the $j$ where $b(n,p,j)$ is largest, the idea is to study the ratio of successive terms as a function of $j$: $$\frac{ b(n,p,j)}{b(n,p,j-1)}=\frac{(n-j+1)}j\frac pq$$ Observe that the ratio decreases as $j$ ranges over $1, 2, \ldots$. As long as the ratio is greater than or equal to $1$, we know that $b(n,p,j)\ge b(n,p,j-1)$, i.e., $b(n,p,j)$ is still increasing.
Case 1: Suppose the value at $j=1$ of this ratio is less than $1$. This occurs when $p(n+1)<1$. Then the ratio never exceeds one, which means $b(n,p,j)$ is decreasing for all $j$ and is maximized at $j^*=0$. Note that we can write $j^*=[p(n+1)]$.
Case 2: Suppose the initial value of the ratio is at least $1$. Then the last time that this ratio is at least one tells you where the maximizing $j$ is. In view of the equivalence $$\frac{(n-j+1)}{j}\frac{p}{q} \ge 1 \iff j\le p(n+1), $$ the last $j$ where the ratio is at least one is $j^* := [p(n+1)]$, the integer part of $p(n+1)$. In the special case where $p(n+1)$ is itself an integer, this translates into the ratio being one at $j^*$, which means $b(n, p, j^*)$ is equal to $b(n,p,j^*-1)$ so $b(n, p, j)$ is maximized also at $j^*-1$.