Showing that ∫ X log ⁡ ( f ) d μ ≤ μ ( X...

Showing that ${\int }_X\log (f)d\mu \le \mu (X)\log \frac{1}{\mu (X)}$ given that ${\int }_{X}fd\mu =1$ for positive $f$ without using Jensen's inequality.
We weren't presented with Jensen's inequality, but with Fubini's Theorem and another theorem which is:
in a finite measure space $f\in L_0^{+}(\mathrm{\Sigma })$ satisfies ${\int }_{X}fd\mu ={\int }_{(0,\mathrm{\infty })}\mu (\{f>t\}dm(t))$
At first I thought I should use $\{|f|<\frac{1}{\mu (X)}\}$,$\{|f|\ge \frac{1}{\mu (X)}\}$ and show that if $\int \log fd\mu >\mu (X)\log (\frac{1}{\mu (X)})$, then
$\int \log fd\mu \le \mu (\{|f|<\frac{1}{\mu (X)}\})\log \frac{1}{\mu (X)}+{\int }_{\{|f|\ge \frac{1}{\mu (X)}\}}\log fd\mu$ and therefore
$\log \frac{1}{\mu (X)}(\mu (X)-\mu (\{|f|<\frac{1}{\mu (X)}\}))=\log \frac{1}{\mu (X)}(\mu (\{|f|\ge \frac{1}{\mu (X)}\}))<{\int }_{\mu (\{|f|\ge \})}\log fd\mu$
which I can't seem to contradict. Any ideas how this can be shown, without using Jensen's Inequality?