Let's start by picking two positive real numbers a,b and see if we can find a third value c such that a+b+c=abc. We'll try to rearrange this to solve for c. Subtract from both sides to get a+b=c(ab−1). If ab>1, we can divide by ab−1 to get the positive value c=a+bab−1. This gives you a solution for any two values a,b such that ab>1.

For ab≤1abc≤c<a+b+c so there are no solutions. This means that we've characterized all solutions in positive real numbers.

Hope I helped you! :)
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