= Solution
With one <edge> colour, $f(1)=3$. Suppose the assertion is known for $k-1$ colours, and take $n=k(f(k-1)-1)+2$. At any <vertex>, among its $n-1$ incident <edges> some colour occurs on at least $f(k-1)$ <edges>. Let $S$ be their other endpoints. If an <edge> within $S$ has that colour, it completes a monochromatic triangle with the chosen <vertex>. Otherwise $S$ uses only $k-1$ colours internally, and the inductive hypothesis gives a monochromatic triangle there. Thus
$$
f(k)\le k(f(k-1)-1)+2.
$$
Induction now yields
$$
f(k)\le k(3(k-1)!-1)+2=3k!-k+2\le3k!\qquad(k\ge2).
$$
Together with the base case, \b[$\boxed{f(k)\le3k!}$ for every positive $k$]. This is a constructive <multicolour Ramsey bound>.
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