Solving an inequality between 2 functions with 2 variables

Question

I have the inequality $$ \frac{p(1-p)}{n} > \frac{(1-n)p^2+(n-2)p+1}{(n+1)^2} $$ where $p \in [0,1]$ and $n$ can be any positive integer. Using Wolfram Alpha I can see that for $n>0$, we have $$ \frac{n}{3n+1} < p < 1 $$ however, I cannot seem to get to this conclusion. I currently do the following $$ \begin{align*} \frac{p(1-p)}{n} &> \frac{(1-n)p^2+(n-2)p+1}{(n+1)^2} \\ p(1-p)(n+1)^2 &> [(1-n)p^2+(n-2)p+1]n \\ 0 &> [(1-n)p^2+(n-2)p+1]n - p(1-p)(n+1)^2 \\ 0 &> n(1-n)p^2+n(n-2)p+n - (p-p^2)(n^2+2n+1) \\ 0 &> (n-n^2)p^2 + (n^2-2n)p + n - n^2p + 2np + 1 - n^2p^2+2np^2+p^2 \\ 0 &> np^2 - n^2p^2 + n^2p - 2np + n - n^2p + 2np + 1 - n^2p^2+2np^2+p^2 \\ 0 &> np + n + 2np^2 + p^2 \\ 0 &> (2n-1)p^2 + np + n \end{align*} $$ and I'm not sure how to proceed from here.

Answer 1

We want $p(1-p)(n+1)^2 - n[(1-n)p^2+ (n-2) p + 1 ] > 0$.

View this as a quadratic in $p$. Expanding and combining terms, we get $$ (3n+1)p^2 - (4n+1) p + n < 0.$$

Factoring, we get $ [ (3n+1) p - n ] [ p - 1 ] < 0$.

Hence $ \frac{n}{3n+1} < p < 1 $.

Answer 2

It's $$(n-n^2+n^2+2n+1)p^2+(n^2-2n-n^2-2n-1)p+n<0$$ or $$(3n+1)p^2-(4n+1)p+n<0.$$ Can you end it now?