Compare $\int_0^1 f(x)\log f(x)\,dx$ and $\int_0^1f(s)\,ds\cdot\int_0^1\log f(t)\,dt$ 
The question: Given $f$ to be a positive, measurable function on $[0,1]$, which is larger, $\displaystyle\int_0^1 f(x)\log f(x)\,dx$ or $\displaystyle\left(\int_0^1f(s)\,ds\right)\left(\int_0^1\log f(t)\,dt\right)$?

I know from testing with $f(x)=x$ that the first integral is, indeed, larger. Of course, this isn't a rigorous "proof" (if it can even be called that). I am unsure as to go about this. At first glance, it looks similar to a Holder's Inequality problem, but appears to go the wrong direction. If anyone has a hint/suggestion as to where to start, it would be much appreciated. Thanks in advance!
EDIT:
I've now found examples which show equality and inequality the other direction. If I let $f(x)=\cases{c\text{ if }x\in[0,1/2)\newline1\text{ otherwise}}$, then the first integral is smaller if $0<c<1$, they are equal if $c=1$, and the second is smaller if $c>1$. The new question, is this a sufficient answer: The magnitude of the integral depends on the function $f$?
RE-EDIT:
I forgot about the sign change of $\log c$ when $c<1$, so I either have the integrals being equal when $c=1$ or the second integral being larger for other values of $c$. I suppose my original question still remains.
 A: This is a particular case of a vastly more general result. To wit, consider two functions $G$ and $H$, both increasing, such that $G(f)$ and $H(f)$ are both integrable. Then, 

$$\int_0^1G(f(x))\mathrm dx\cdot\int_0^1H(f(x))\mathrm dx\leqslant\int_0^1G(f(x))H(f(x))\mathrm dx.$$

To show this, integrate on $(0,1)\times(0,1)$ with respect to the Lebesgue measure the inequality $$[G(f(x))-G(f(y))]\cdot[H(f(x))-H(f(y))]\geqslant0,$$ which holds everywhere in $(0,1)\times(0,1)$, and expand the inequality you get.
Thus, the result in the question uses only the monotonicity of the functions $t\mapsto t$ and $t\mapsto\log t$.
A: Let me give you a sequence of hints:

 1. Do you know Jensen's inequality?

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 2. $\frac{d^2}{dx^2} (x \cdot \ln(x)) = \frac{d}{dx} (\ln(x) + x \cdot 1/x) = 1/x > 0$ for $x>0$. Hence, the map $\Phi : (0,\infty) \to \Bbb{R}, x \mapsto x \cdot \ln(x)$ is convex.

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 3. An application of Jensen's inequality yields $$\int_0^1 f(x) dx \cdot \ln \left(\int_0^1 f(x) dx\right) =\Phi \left( \int_0^1 f(x)\,dx\right)\leq \int_0^1 \Phi(f(x)) \, dx$$

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 4. Finally, $\frac{d^2}{dx^2}(-\ln(x)) = \frac{d}{dx} -1/x = x^{-2} > 0$, so that $-\ln : (0,\infty) \to \Bbb{R}$ is also convex. Another application of Jensen gives $$- \ln\left(\int_0^1 f(x) \, dx \right) \leq \int_0^1 -\ln(f(x))\,dx$$. Together with step (3) and with $\int_0^1 f(x) \,dx > 0$, this proves the claim.

