A simple quadratic inequality For positive integers $n\ge c\ge 5$, why does
$$c+2(n-c)+\frac{(n-c)^2}{4}\le\frac{(n-1)^2}{4}+1\text{ ?}$$
 A: Essentially, you want to prove that (multiplying throughout by $4$) $$c^2 -2cn -4c +n^2 + 8n \leq n^2 -2n +5$$
i.e. $$c^2 - 2cn -4c + 10n \leq 5$$
Since $n \geq c \geq 5$, we have that $(10-2c)n \leq (10-2c)c$ (since $10-2c \leq 0$). Hence, $$c^2 -4c + (10-2c)n \leq c^2 - 4c +(10-2c)c = -c^2 + 6c = c(6-c) = 9 - (c-3)^2$$ Note that the function is a decreasing function for $c \geq 3$. Hence, it is a decreasing function for $c \geq 5$ as well. Hence, the maximum occurs at $c=5$, which gives us that $c(6-c) \leq 5 (6-5) = 5$, which is what we wanted.
Putting these together, we get that $$c+2(n-c)+\frac{(n-c)^2}{4}\le\frac{(n-1)^2}{4}+1$$
A: 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. 
A: If you expand the two sides, you will get
\begin{equation}
\frac{c^2}{4}-\frac{cn}{2}-c+\frac{n^2}{4}+2n\leq\frac{n^2}{4}-\frac{n}{2}+\frac{5}{4}
\end{equation}
\begin{equation}
\frac{c^2}{4}-\frac{5}{4}-c\leq n\frac{c-5}{2}
\end{equation}
If c=5 than the inequality is satisfied, assuming $5\leq c$ we have
\begin{equation}
\frac{c-5}{2} -\frac{2c}{c-5}\leq n
\end{equation}
I think you can use induction to prove this inequality.
