Let $a$, $b$, and $c$ be non-zero real number such that $\tfrac{ab}{a+b}=3$, $\tfrac{bc}{b+c}=4$, and $\tfrac{ca}{c+a}=5$. There are relatively prime positive integers $m$ and $n$ so that $\tfrac{abc}{ab+bc+ca}=\tfrac{m}{n}$. Find $m+n$.
1. We start with the given equations:
   \[
   \frac{ab}{a+b} = 3, \quad \frac{bc}{b+c} = 4, \quad \frac{ca}{c+a} = 5
   \]
   These can be rewritten in terms of reciprocals:
   \[
   \frac{1}{a} + \frac{1}{b} = \frac{1}{3}, \quad \frac{1}{b} + \frac{1}{c} = \frac{1}{4}, \quad \frac{1}{c} + \frac{1}{a} = \frac{1}{5}
   \]

2. Add the three equations:
   \[
   \left( \frac{1}{a} + \frac{1}{b} \right) + \left( \frac{1}{b} + \frac{1}{c} \right) + \left( \frac{1}{c} + \frac{1}{a} \right) = \frac{1}{3} + \frac{1}{4} + \frac{1}{5}
   \]
   Simplify the left-hand side:
   \[
   2 \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)
   \]
   Simplify the right-hand side:
   \[
   \frac{1}{3} + \frac{1}{4} + \frac{1}{5} = \frac{20}{60} + \frac{15}{60} + \frac{12}{60} = \frac{47}{60}
   \]

3. Equate the simplified expressions:
   \[
   2 \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) = \frac{47}{60}
   \]
   Divide both sides by 2:
   \[
   \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{47}{120}
   \]

4. We need to find the value of \(\frac{abc}{ab + bc + ca}\). Using the identity:
   \[
   \frac{abc}{ab + bc + ca} = \frac{1}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}
   \]
   Substitute \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{47}{120}\):
   \[
   \frac{abc}{ab + bc + ca} = \frac{1}{\frac{47}{120}} = \frac{120}{47}
   \]

5. The fraction \(\frac{120}{47}\) is already in its simplest form since 120 and 47 are relatively prime.

6. Therefore, \(m = 120\) and \(n = 47\). The sum \(m + n\) is:
   \[
   m + n = 120 + 47 = 167
   \]

The final answer is \(\boxed{167}\)