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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.

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    You need to show that $mn\geq m+n-1$, or equivalently $m(n-1)\geq n-1$. – 2017-02-20
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    is here $$m,n\geq 1$$? – 2017-02-20
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    Starting from the second expression, Consider the case where $n = 1$ then equality holds. Consider any other value for $n$ then we can safely divide by $(n-1)$. We get $m > 1$ or $m=1$. which is true given that they are both positive integers. Then, multiplying we get the first expression which I want to prove.. @LuizCordeiro – 2017-02-20
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    @Dr.SonnhardGraubner They are positive integers. Doesn't that imply they are both larger or equal to one? – 2017-02-20

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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 $$m,n\geq 1$$

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    This is simple and beautiful! Thank you! – 2017-02-20
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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 $n\geq1$

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    Couldn't think of induction. Thanks for your answer! – 2017-02-20