Use the weighted mean value theorem (Theorem 3.16 in Apostol) to prove:
Recall the weighted mean value theorem:
For functions and continuous on , if never changes sign in then there exists such that
Then substituting our definitions of and ,
Since and are continuous and does not change sign on we may apply Theorem 3.16,
Since is strictly decreasing on , we have
It seems safer to do integration on the product of g and f, as in the proof of 3.16 (and not like done here).
Sorry, the way you proved it seems valid (apart from the mentioned error in the last row) and probably the intended way, because it uses the required theorem.
The inequality you end up with is not the one the question is asking for; there is an extra factor of f(c).
remember that f(c)(integral of g(x)) = (integral of f(x)g(x)) which is the original equation. In other words both the function he ended up with and the one being asked for are equivalent.
*edit although he did make a typo and rewrote f(c) in the final line when he had already factored it back in.