From the binomial expansion: c_r=\binom{n}{r}a^{n-r}b^r=\frac{n!}{r!(n-r)!}a^{n-r}b^r
Assuming m\not=0, c_m\geq c_{m-1}:
\begin{align}
c_m&\geq c_{m-1}\\
\frac{n!}{m!(n-m)!}a^{n-m}b^m&\geq\frac{n!}{(m-1)!(n-m+1)!}a^{n-m+1}b^{m-1}\\
b(n-m+1)&\geq am\\
b(n+1)&\geq am+bm\\
\frac{b(n+1)}{a+b}&\geq m\\
\end{align}
Since n,a,b>0, \frac{b(n+1)}{a+b}>0, so the result also holds for m=0.
Similarly, assuming m\not=n:
\begin{align}
c_m&\geq c_{m+1}\\
\frac{n!}{m!(n-m)!}a^{n-m}b^m&\geq\frac{n!}{(m+1)!(n-m-1)!}a^{n-m-1}b^{m+1}\\
a(m+1)&\geq b(n-m)\\
am+bm&\geq bn-a\\
m&\geq \frac{bn-a}{a+b}\\
m&\geq \frac{bn-a}{a+b}+1-1\\
m&\geq \frac{bn-a}{a+b}+\frac{a+b}{a+b}-1\\
m&\geq \frac{b(n+1)}{a+b}-1\\
\end{align}
For the special case n=m, we first note that \frac{b}{a+b}<1:
\begin{align}
1&\geq \frac{b}{a+b}\\
n+1&\geq \frac{b(n+1)}{a+b}\\
n&\geq \frac{b(n+1)}{a+b}-1\\
m&\geq \frac{b(n+1)}{a+b}-1\\
\end{align}
so the result still holds.
Combining these two inequalities gives the stem result.
Since k-1\leq m\leq k for some k, m can take at most 2 integral values.
For the next parts, we start by noting that G(n,a,b)=\left\lfloor\frac{b(n+1)}{a+b}\right\rfloor.
part (i)
\begin{align}
G(9,1,3)&=\left\lfloor\frac{3(9+1)}{1+3}\right\rfloor=7\\
G(9,2,3)&=\left\lfloor\frac{3(9+1)}{2+3}\right\rfloor=6
\end{align}
part (ii)
\begin{align}
G(2k,a,a)&=\left\lfloor\frac{a(2k+1)}{a+a}\right\rfloor\\
&=\left\lfloor\frac{2k+1}2\right\rfloor\\
&=k\\\\
G(2k-1,a,a)&=\left\lfloor\frac{a(2k-1+1)}{a+a}\right\rfloor\\
&=\left\lfloor\frac{2k}2\right\rfloor\\
&=k
\end{align}
part (iii)
In order to maximise G(n,a,b)=\left\lfloor\frac{b(n+1)}{a+b}\right\rfloor, we want to minimise the denominator. This is achieved when a=1, since a is a positive integer.
part (iv)
Using the fact that \frac{b}{1+b}<1:
\begin{align}
G(n,1,b)&=\left\lfloor\frac{b(n+1)}{1+b}\right\rfloor\\
&=\left\lfloor\frac{b}{1+b}(n+1)\right\rfloor\\
&<n+1
\end{align}
Therefore, G(n,1,b) is at most n. This is achieved when:
\begin{align}
\left\lfloor\frac{b(n+1)}{1+b}\right\rfloor&=n\\
\frac{b(n+1)}{1+b}&\geq n\\
b(n+1)&\geq n(1+b)\\
b&\geq n
\end{align}
