I have this case of binomial theorem: $$ \sum_{i=0}^n \binom{n}{i} r^i (n-r)^{n-i}. $$
Now, for some reason, we know that $r\le \frac{n}{2}$. Is this enough to conclude that $$ \sum_{i=0}^n \binom{n}{i} r^i (n-r)^{n-i} > \sum_{i=0}^r \binom{n}{i} r^r (n-r)^{n-r} ? $$
I assume you mean $0 < r \le n/2$. Then $r \le n-r$, so for $0 \le i \le r$ you have $$ r^{i-r} \ge (n-r)^{i-r}$$ and thus $$ r^i (n-r)^{n-i} \ge r^r (n-r)^{n-r}$$ Using this for $i=0$ to $r$, and putting additional positive terms on the left for $i=r+1$ to $n$, we do indeed have $$ \sum_{i=0}^n {n \choose i} r^i (n-r)^{n-i} > \sum_{i=0}^r {n \choose i} r^r (n-r)^{n-r} $$