A minimum value proof

Question

Let $m$ and $n$ be positive integers and let $x = m + n$. I want to prove that the minimum value of $m$ $*$ $n$ is equal to $x-1$. It happens when one of $m$ or $n$ is equal to one.

I tried to write the multiplication as a quadratic equation $(x-m)(m) = -m^2 + xm = -(m -0.5n)^2 + 0.25n^2$, But that didn't help with the minimum.

Answer 1

we want to prove that $$mn\geq m+n-1$$ then our statement is equívalent to $$(m-1)(n-1)\geq 0$$ which is clearly true, if $$m,n\geq 1$$

Answer 2

How about induction in $m$ with $n$ fixed $n\geq1$. In case $m=1$ is obvious. Then if $mn\geq m+n-1$, then $$\begin{align}(m+1)n&=mn+n\\&\geq m+n-1+n\\&=((m+1)+n-1)+(n-1)\\&\geq (m+1)+n-1\end{align}$$ as $n\geq1$