Prove that $\int_E \log fd\mu \leqslant \mu(E) \, \log \left[\frac{1}{\mu(E)} \right]$ for strictly positive measure $\mu$
Define
$$\mu_E(dy) := \frac{1}{\mu(E)} 1_E(y) \, d\mu(y)$$
then
$$\int_E \log f \, d\mu = \mu(E) \int \log f \, d\mu_E.$$
Applying Jensen's inequality to the probability measure $\mu_E$ and $F(x) = \log x$ we find
$$\int_E \log f \, d\mu \leq \mu(E) \log \left( \int f \, d\mu_E \right) = \mu(E) \log \left( \frac{1}{\mu(E)} \int_E f \, d\mu \right).$$
As $\int_E f \, d\mu \leq 1$ it follows from the monotonicity of the logarithm that
$$\int_E \log f \, d\mu \leq \mu(E) \log \left( \frac{1}{\mu(E)} \right). $$