Well my argument is sufficient to show that a solution where two adjacent filters don't follow this order can't be optimal, because otherwise swapping those adjacent filters f[i] and f[i+1] would improve the result.

So any optimal solution must follow this strict order criteria. ¹

qed! =)

PS: I'm glad I didn't start implementing the B&B algorithm :-)

UPDATE:

¹) and it's easy to see that all ordered solutions (plural b/c adjacent filters can have the same weight) imply the same total cost.