Easy to prove that $0<r<3$ Let $b=c=1, a=\sqrt{\frac{3-r}{2r}}$ We have $r(2a+1)+(3-r) (\frac{1}{a}+2)= 2ar+\frac{3-r}{a} +6-r = 2\sqrt{2r(3-r)}+6-r \leq (2r+3-r)+6-r=9$ with equality for $r=1$ And for $r=1$ we have $ (ab + bc + ca) + 2 \left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right) \ge 9$ because $ab+\frac{1}{a}+\frac{1}{b} \geq 3$

splendid solution!

teomihai wrote: splendid solution! indeed