TITLE: Measurable function on the interval $[0,1]$
QUESTION [8 upvotes]: Assume that $f$ is a measurable function on the interval $[0,1]$ such that $0<f(x)<\infty$ for $x \in [0,1]$. Then, how can I prove the inequality below?
$$\int^1_0 f(x)dx \int^1_0 {1 \over {f(x)}} dx \ge 1$$

REPLY [2 votes]: $$\iint_{[0,1]^2}\frac{(f(x)-f(y))^2}{f(x)f(y)}\mathrm dx\mathrm dy\geqslant0$$
Edit: This is an adaptation of the well-known approach to the inequality $\|f\|_1\leqslant\|f\|_2$ for every probability measure $\mu$ as the expansion of
$$
\iint(f(x)-f(y))^2\mathrm d\mu(x)\mathrm d\mu(y)\geqslant0$$
and gives an opportunity to recommend once more the marvelous little book The Cauchy-Schwarz Master Class: An Introduction to the Art of Inequalities by J. Michael Steele.