The uvw method

Schur's inequality. For $r=1$:
$$p^3+9r\ge4pq,$$
i.e. $27+9\ge12q$, so $q\le3$.

Conclusion. $ab+bc+ca\le3$, with equality iff $a=b=c=1$.

Alternative proof. Since $a,b,c$ are roots of $X^3-pX^2+qX-r=0$, the condition $abc=1$ and $a+b+c=3$ constrains the discriminant. The discriminant $p^2q^2-4q^3-4p^3r-27r^2+18pqr\ge0$ becomes $81q^2-4q^3-108-27+54q\ge0$. Combined with $q\le3$, this is consistent.

Conclusion. The result follows from Schur's inequality or discriminant analysis.