Showing that $\int_{X}\log(f)d\mu\le \mu(X)\log{1\over \mu (X)}$ without using Jensen's inequality
Solution 1:
Use that $$\log(x) \leq \log(s) - 1 + \frac xs$$ for any constant $s >0$. Then take $s= \mu(X)^{-1}$ and use this bound to get an upperbound for the integrand and hence also for the integral. (This is, all said and done, just Jensen’s inequality.)