A simple quadratic inequality For positive integers $n\ge c\ge 5$, why does
$$c+2(n-c)+\frac{(n-c)^2}{4}\le\frac{(n-1)^2}{4}+1\text{ ?}$$

To avoid fractions, we multiply the left-hand side by $4$, obtaining 
$$(n-c)^2+8(n-c)+4c.$$
Complete the square. We get 
$$(n-c+4)^2 +4c -16.$$
Now calculate $[(n-1)^2 +4]-[(n-c+4)^2 +4c -16]$.
The difference of squares factors as $(c-5)(2n-c+3)$, so
$$\begin{align}[(n-1)^2 +4]-[(n-c+4)^2 +4c -16]&=(c-5)(2n-c+3)-4(c-5)\\
&=(c-5)(2n-c-1).\end{align}$$
The condition $c \ge 5$ ensures that $c-5\ge 0$. And since $n \ge c$, the term $2n-c-1$ is positive. 
