A simple quadratic inequality

Question

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

Answer 1

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.

Answer 2

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 $$c+2(n-c)+\frac{(n-c)^2}{4}\le\frac{(n-1)^2}{4}+1$$

Answer 3

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.