Define $$g_{(k,j)} = \frac{a^{n-k}b^k(k+n)!x^{k+n-j}}{k!(n-k)!(k+n-j)!}$$, where $n,k,j \in \Bbb{N}$ are fixed such that $(0 \leq x \leq a/b ),(b<a),(0 \leq k \leq n ),(2 \leq j \leq 2n),(0 \leq k+n-j)$.
By the continuity of $g_{(k,j)}$ in compact $[0,a/b]$, there exists $x_{(k,j)} \in [0,a/b]$ that is the maximum of $g_{(k,j)}(x)$.
Define $A = \{(k,j) : (0 \leq k \leq n) and (2 \leq j \leq 2n)\}$ that is finite, so there exists $x_{m} \in [0,a/b]$, and $(k_{m},j_{m})$ such that
$g_{(k_{m},j_{m})} (x_{m}) = max_{(k,j) \in A}\{g_{(k,j)}(x_{(k,j)}): x_{(k,j)}$ is the maximum of $g_{(k,j)}(x)\}$
We are interested in find $(k,j,x)$ that is $g_{(k_{m},j_{m})} (x_{m})$.
It is maximized when $x$ is as large as possible, so $x=a/b$.
The ratio $g(k,j+1)/g(k,j)=(b/a)(k+n-j)$. So long as the ratio is more than 1, $g(k,j)$ is increasing with $j$, but when the ratio dips below 1, $g(k,j)$ starts decreasing. so it is maximized when $j=k+n-a/b$ - or else at an endpoint $j=2$ or $j=k+n$