An inequality involving factorials and two variables

Question

The problem is as follows:

For $m\ge n>1$ prove that $$(m-2)!(n-1)+(n-2)!(m-1)+(m-2)(n-2)\ge (m-1)(n-1)$$

Since $(m-1)(n-1)-(m-2)(n-2)=m+n-3$ so we only need to show that $$(m-2)!(n-1)+(n-2)!(m-1)\ge m+n-3$$. On the face of it this seems to hold but what would be a rigorous way of showing this?

Answer 1

Hint: Since $m\ge n$, write $m=n+k$ with $k\ge 0$ and do induction over $k$. (So actually the induction is over $m-n$). For $k=0$ it is obvious (but let's do it also for $k=1$). So,

1. Base case: $k=1$ \begin{align} (n-1)!(n-1)+(n-2)!n+(n-1)(n-2)&\ge n(n-1) \iff\\ (n-1)!+\frac{(n-2)!n}{n-1}+n-2&\ge n\iff\\ (n-1)!+\frac{(n-1)!n}{(n-1)^2}&\ge 2\iff\\ (n-1)!\left(1+\frac{n}{(n-1)^2}\right)&\ge 2\end{align} which is obviously true for $n\ge 2$ (for $n=2$ you can check it directly and for $n\ge 3$: $(n-1)!\ge2$ and the term in parenthesis is $>1$).
2. Induction step: $m=n+k \to m=n+k+1$. I will use your expression:

\begin{align}m+n-3&=(n+k+1)+n-3\\&=\underbrace{(n+k)+n-3}_{\text{Induction hypothesis}}+1\\&\le (n+k-2)!(n-1)+(n-2)!(n+k-1)+1\\&=(n+k-2)!(n-1)+(n-2)!(n+k)-(n-2)!+1 \\&\le (n+k-2)!(n-1)+(n-2)!(n+k) \\&\le (n+k-1)!(n-1)+(n-2)!(n+k)\\&=(m-2)!(n-1)+(n-2)!(m-1)\end{align}