Show $\int_0^1 f(x) dx \int_0^1 \ln f(x) dx \le \int_0^1 f(x) \ln f(x) dx$, when $f \ge 1$ Let $f:[0,1] \to [1,\infty)$ be such that
$$
\int_0^1f(x)\ln f(x) dx < \infty. \tag{1}
$$
Show
$$
\int_0^1 f(x) dx \int_0^1 \ln f(x) dx \le \int_0^1 f(x) \ln f(x) dx. \tag{2}
$$
From (1), $f$ and $\ln f$ are integrable on $[0,1]$.
Using this post, $\ln x \le x \ln x$, when $x >0$.
I can't go further.
 A: Here's a cute trick I learned in a probability course. You can apply Fubini's theorem to show that the inequality stated is equivalent to the positivity of the following double integral:
$ \int_{[0,1]^2} \left( f(x) - f(y) \right) \left(\ln f(x) - \ln f(y) \right) dx dy$
That's the cute part. Notice that there's a kind of 'evenness' here for us to exploit: decompose [0,1]^2 as $\{(x,y) : f(x) \geq f(y)\} =: A$ and its complement. Now on $A$ the integrand is positive by the monotonicity of $\ln$ and the assumption $f \geq 1$. On $A^c$ the integrand is the product of two negatives, hence also positive. Therefore the integral is positive and the inequality holds.
I'll also mention that the above procedure can be applied to a very wide variety of situations (but in order to apply it you must have a condition analogous to (1) to use Fubini!).
A: $g(x):=x\log(x)$ is convex: $g'(x)=\log(x)+1$, $g''(x)= \frac{1}{x}\geq 0$, the second derivative implying convexity. Then, by Jenson's inequality:
$\int f(x)\ln f(x)dx \geq g(\int_0^1f(x)dx)=\left(\int_0^1f(x)dx\right) \log(\int_0^1f(x)dx)\geq \int_0^1f(x)dx\int_0^1\log f(x)dx$
where in the last step we use concavity of $\log(x)$ and Jenson's inequality again for concave functions this time.
