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

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This inequality is wrong. –  Did Jun 18 '12 at 18:35
It comes from the third page of the paper ftp.inf.fu-berlin.de/pub/math/publ/pre/1995/pr-a-95-24.ps (the link looks currently down though) –  Fred Jun 18 '12 at 18:42
...Where I am sure the additional condition $c\geqslant5$ is not omitted--unlike in the first version of the question here. –  Did Jun 18 '12 at 19:35
@did Yes, sorry, the original post contained this mistake. –  Fred Jun 25 '12 at 14:56

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.

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But the last assertion of your answer is wrong: we have $c\ge 5$, not $c\le 5$. –  Fred Jun 25 '12 at 14:43
@Fred: Thank you. I had a typo on the last line. The condition $c\ge 5$ makes $4c-16\ge 4$. Which is what we want. –  André Nicolas Jun 25 '12 at 14:52
But we need to show that $((n-c)+4)^2+4c-16\le (n+1)^2+4$, so $4c-16\ge 4$ is the wrong direction, no ? –  Fred Jun 25 '12 at 14:53
Thanks for your help! –  Fred Jun 25 '12 at 15:54
@Fred: The thanks go the other way, you helped fix an error. –  André Nicolas Jun 25 '12 at 16:00

If you expand the two sides, you will get

$$\frac{c^2}{4}-\frac{cn}{2}-c+\frac{n^2}{4}+2n\leq\frac{n^2}{4}-\frac{n}{2}+\frac{5}{4}$$

$$\frac{c^2}{4}-\frac{5}{4}-c\leq n\frac{c-5}{2}$$

If c=5 than the inequality is satisfied, assuming $5\leq c$ we have

$$\frac{c-5}{2} -\frac{2c}{c-5}\leq n$$

I think you can use induction to prove this inequality.

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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 $$c+2(n-c)+\frac{(n-c)^2}{4}\le\frac{(n-1)^2}{4}+1$$
Alternatively,$$(n-c)^2\leq (n-5)^2 \leq (n-5)^2 + (4c-20)$$ Multiplying out and subtracting gives your $c^2-2cn-4c+10n\leq 5$ as you've already shown is required. –  Thomas Andrews Jun 18 '12 at 18:56