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

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