For all $m\geq 3$ and $n\geq 1$, show that $\big(\frac{mn^2-mn+2}{2}\big) ^2-\big(\frac{m(n-1) ^2-m(n-1) +2}{2}\big) \big(\frac{m(n+1) ^2-m(n+1) +2}{2}\big)\geq 0$.
I simplified the expression and got $\frac{m^2n^2-m^2n-2m}{2}$ which I couldn't justified its non-negativity. Can anyone suggest me other options to easily show the non-negativity of the expression, please?
We can rewrite it more compactly as follows: $$ \begin{split} &\phantom{=}\left(m\binom{n}{2}+1\right)^2-\left(m\binom{n-1}{2}+1\right)\left(m\binom{n+1}{2}+1\right)\\ &=m^2\left(\binom{n}{2}^2-\binom{n-1}{2}\binom{n+1}{2}\right)-m\left(\binom{n+1}{2}-2\binom{n}{2}+\binom{n-1}{2}\right)\\ &=m^2\binom{n}{2}-m\binom{n-1}{0}=m^2\binom{n}{2}-m, \end{split} $$ so it is nonnegative for $n\ge 2$ and any integer $m$. But for $n=1$, it is equal to $-m$, so it's only nonnegative for $m\le 0$.
On the other if we had "$-1$" instead of "$+1$" in all three parentheses, the result would indeed be nonnegative, i.e. $$ \left(m\binom{n}{2}-1\right)^2-\left(m\binom{n-1}{2}-1\right)\left(m\binom{n+1}{2}-1\right)=m^2\binom{n}{2}+m\ge 0 $$ for all nonnegative $m$ and $n$.