(Feller volume 1, Q28, p.171) Suppose that we have the following binomial distribution $$\frac{n!}{k_1!k_2!... k_r!} p_1^{k_1}p_2^{k_2} ... p_r^{k_r}.$$ Prove the theorem. The maximal term of the multinomial distribution satisfies the inequalities $$np_i -1 < k_i \le (n+r-1) p_i,$$ for $i=1, ... , r$. Hint: Prove first that the term is maximal if, and only if , $p_i k_j \le p_j(k_i+1)$ for each pair $(i,j)$. Add these inequalities for all $j$, and also for all $i \not= j$.
I first do not know how to prove the first hint. Assuming that we have the inequality in the hint, $\sum_{i \not= j} \sum_j p_ik_j \le \sum_{i \not= j} \sum_j p_j(k_i +1) \implies n(1-p_j) \le n-k_j+ r-1$. This is all I can think now. I really appreciate if you give some help.
For the maximum term we should have for each pair $(i,j), i \neq j$: $$ \frac {\frac{p_i^{k_i} p_j^{k_j}} {k_i! k_j!}} {\frac{p_i^{k_i+1} p_j^{k_j-1}} {(k_i+1)!(k_j-1)!} } \geq 1, $$
which gives $$ \frac{(k_i + 1)}{k_j} \cdot \frac{p_j}{p_i} \geq 1, $$ $$ (k_i + 1)p_j \geq k_j p_i \tag{*}\label{*}$$ and the hint is proved.
Now adding both sides of \eqref{*} for all $j$ we get $ \sum_j k_j p_i = n p_i $ and $ \sum_j p_j (k_i + 1) = k_i + 1, $ and since $\sum_i k_i = n$ and $\sum_i (n p_i - 1) < n $ we get $$ n p_i - 1 < k_i. \tag{**}\label{**} $$
Adding both sides of \eqref{*} for all $i \neq j$ we obtain $$ \sum_{i \neq j} k_j p_i = k_j \left(\sum_i (p_i) - p_j\right) = k_j - k_j p_j, $$ $$ \begin{align} \sum_{i \neq j} p_j (k_i + 1) & = p_j\left(\sum_i k_i - k_j\right) + p_j(r-1) = \\ & = p_j(n-k_j) + p_j(r-1) = \\ & = p_j(n+r-1) - p_j k_j, \end{align} $$
and now
$$ k_j - k_j p_j \leq p_j(n+r-1) - p_j k_j, $$ or $$ k_j \leq p_j(n+r-1), $$ where we are allowed to replace $j$ with $i$ to obtain $$ k_i \leq p_i(n+r-1), \tag{***}\label{***} $$
and combining $\eqref{**}$ and \eqref{***} we finally get Feller's (10.1).