Does existence of a total $\mathsf{NP}$ search problem not solvable in polytime imply $\mathsf{NP}\cap\mathsf{coNP} \neq \mathsf{P}$? It easy to see that if $\mathsf{NP}\cap\mathsf{coNP} \neq \mathsf{P}$ then there are total $\mathsf{NP}$ search problems which cannot be solved in polynomial time (create a total search problem by having both the witnesses for membership and the witnesses for nonmembership).

Is the converse also true, i.e. 


  Does existence of a total $\mathsf{NP}$ search problem not solvable in polynomial time imply $\mathsf{NP}\cap\mathsf{coNP} \neq \mathsf{P}$?
For total UP the answer is yes, because the question "Is the i'th bit of the solution 1?" is in $NP \cap co-NP$.