Measurable function on the interval $[0,1]$ 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?
$$\int^1_0 f(x)dx \int^1_0 {1 \over {f(x)}} dx \ge 1$$
 A: Here is another way. What you have written is nothing but the Arithmetic mean-Harmonic mean inequality for functions. In case, of a finite set of positive numbers $\{a_i\}$'s, the AM-HM inequality reads
$$\dfrac{\sum_{i=1}^{n} w_i a_i}{\sum_{i=1}^{n} w_i} \geq \dfrac{\sum_{i=1}^{n} w_i}{\sum_{i=1}^{n} \dfrac{w_i}{a_i}}$$where $w_i \geq 0$ are the weights. The weights can be normalized to $1$ i.e. if we enforce that $\displaystyle \sum_{i=1}^{n} w_i = 1$, then we get that $$\sum_{i=1}^{n} w_i a_i \geq \dfrac1{\displaystyle \sum_{i=1}^{n} \dfrac{w_i}{a_i}}$$
Hence, we get that $$\left(\sum_{i=1}^{n} w_i a_i \right) \left(\sum_{i=1}^{n} \dfrac{w_i}{a_i} \right)\geq 1$$
If you partition the interval $[0,1]$ into $n$ disjoint measurable sets, say $E_1^{(n)},E_2^{(n)},\ldots,E_n^{(n)}$, such that $\displaystyle \bigcup_{k=1}^{n} E_k^{(n)} = [0,1]$, choose $w_i^{(n)} = \mu^{(n)} \left(E_i^{(n)}\right)$. Note that $\sum_i w_i^{(n)} = 1$.
Approximate $f(x)$ from below using step functions on these intervals i.e. $f_{step}^{(n)} = f_i^{(n)}$ on the interval $E_i^{(n)}$ such that $\displaystyle \sum_{i=1}^{n} f_i^{(n)} \mu(E_i^{(n)}) = \int f_{step}^{(n)} \to \int f$.
Since $f>0$ on $[0,1]$, $\dfrac1f$ can be approximated from above using the step function $(1/f)_{step}^{(n)} = 1/f_i^{(n)}$ on the interval $E_i^{(n)}$ such that $\displaystyle \sum_{i=1}^{n} \dfrac1{f_i^{(n)}} \mu(E_i^{(n)}) = \int \dfrac1{f_{step}^{(n)}} \to \int \dfrac1{f}$.
From the AM-HM inequality we have that $\left(\displaystyle \sum_{i=1}^{n} \mu(E_i^{(n)}) f_i^{(n)} \right) \left( \displaystyle \sum_{i=1}^{n} \dfrac{\mu(E_i^{(n)}) }{f_i^{(n)}} \right) \geq 1$.
Take the limit to conclude that $$\left(\int f dx \right) \times \left(\int \dfrac1f dx \right)\geq 1$$
You may need to argue out and justify some parts of the above argument to make it into a rigorous proof but hope the idea is clear.
A: $$\iint_{[0,1]^2}\frac{(f(x)-f(y))^2}{f(x)f(y)}\mathrm dx\mathrm dy\geqslant0$$
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.
A: We have by Cauchy-Schwarz inequality,
$$1=\int_0^1 1dx=\int_0^1\sqrt{f(x)}\frac 1{\sqrt{f(x)}}dx\leq \left(\int_0^1f(x)dx\cdot\int_0^1\frac 1{f(x)}dx\right)^{1/2}.$$
(the result is clear if $f^{1/2}$ or $f^{-1/2}$ is not integrable.
A: As @PZZ commented below @David Giraudo's answer, we may assume $f$ is integrable. Then Jensen's inequality, with $\varphi(x)=1/x$ says that
$$
\int_{0}^{1} \frac{1}{f(x)}\, dx = \int_{0}^{1}\varphi(f(x))\,dx \geq \varphi\left[\int_{0}^{1} f(x)\, dx\right] = \left[\int_{0}^{1}f(x)\,dx\right]^{-1}.
$$
