Does $VP \neq VNP$ imply $P \neq NP$?
The short answer is 'no'. No such implication is known. There are two main obstacles: Going from arithmetic circuit complexity to boolean complexity (VP≠VNP implies P/poly≠NP/poly) and then going from boolean circuit complexity (P/poly≠NP/poly) to uniform complexity (P≠NP). Neither of these steps is known. I believe that P/poly≠NP/poly implies VP≠VNP, however.