Big Rudin Exercise 3.26 - Which integral is larger

The function $x\mapsto x\log x$ is convex on $(0,\infty)$, as its second derivative is positive. Thus by Jensen's inequality,

$$\int_0^1 f(t)\log f(t) dt \geq \int_0^1 f(t) dt \log\left( \int_0^1 f(t) dt \right) .$$

The function $x\mapsto \log x$ is concave, so another application of Jensen's inequality yields

$$\log\left( \int_0^1 f(t) dt \right) \geq \int_0^1\log f(t) dt .$$

Combining these two inequalities proves the result.

$$\int_0^1f(t)d(t)=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n f\left(\frac{i}{n}\right)$$ So in order to prove the inequality $$ \int_0^1 f(x) \log f(x) dx \geq \int_0^1f(s)ds \int_0^1\log f(t)dt $$ it is adequate to show $$ \frac{1}{n}\sum_{i=1}^n f\left(\frac{i}{n}\right) \log f\left(\frac{i}{n}\right) \geq \frac{1}{n}\sum_{j=1}^n f\left(\frac{j}{n}\right) \cdot \frac{1}{n}\sum_{k=1}^n\log f\left(\frac{k}{n}\right) $$

Since $\log f(x)$ increases as $f(x)$ increases, we can apply Chebychev's inequality to give $$ \sum_{j=1}^n f\left(\frac{j}{n}\right) \cdot \sum_{k=1}^n\log f\left(\frac{k}{n}\right) \leq n \sum_{i=1}^n f\left(\frac{i}{n}\right) \log f\left(\frac{i}{n}\right) $$ from which the required result follows immediately.