Inequality $\left(\int_0^1 f(x)dx\right) \left(\int_0^1 \frac{1}{f(y)} dy\right) \ge 1$ Let $f$ be a positive continuous function defined on a closed interval $[0,1]$, then it is true that:
$$\left(\int_0^1 f(x)dx\right) \left(\int_0^1 \frac{1}{f(y)} dy\right) \ge 1$$
I tried to show this by definition using Riemann sum, and also by Darboux sums, but it seems too messy, and I am not completely sure that it will lead me to result. How it should be proved?
P.S. Although this problem looks like a homework problem, I am simply curious of how it could be proved.
 A: This follows immediately from the Cauchy–Schwarz inequality:
$$
\left(\int_0^1 f(x)dx\right) \left(\int_0^1 \frac{1}{f(x)} dx\right) \ge \left(\int_0^1 \sqrt{f(x)} \frac{1}{\sqrt{f(x)}} dx\right)^2 = 1
$$
A: Integrate on $[0,1]\times[0,1]$ the inequality
$$
(f(x)-f(y))\cdot\left(\frac1{f(x)}-\frac1{f(y)}\right)\leqslant0.
$$
A: Take $f_1(x) = \sqrt{f(x)}, f_2(x) = {1 \over \sqrt{f(x)}}$. Then Cauchy Schwarz gives $1 = |\langle f_1, f_2 \rangle | \le \|f_1\|_2 \|f_2\|_2$. Squaring gives the desired result.
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
&\color{#ff0000}{\int_{0}^{1}\fermi\pars{x}\,\dd x\int_{0}^{1}{1 \over \fermi\pars{x'}}\,\dd x'}
=
\half
\int_{0}^{1}\int_{0}^{1}\bracks{%
{\fermi\pars{x} \over \fermi\pars{x'}} + {\fermi\pars{x'} \over \fermi\pars{x}}}
\,\dd x\,\dd x'
\\[3mm]&=
\half
\int_{0}^{1}\int_{0}^{1}\braces{\vphantom{\Huge A^{A}}\,\bracks{\vphantom{\LARGE A^{A^{A}}}%
\root{\fermi\pars{x} \over \fermi\pars{x'}}
- \root{\fermi\pars{x'} \over \fermi\pars{x}}}^{\,2} + 2\root{\fermi\pars{x} \over \fermi\pars{x'}}\root{\fermi\pars{x'} \over \fermi\pars{x}}\,}
\,\dd x\,\dd x'
\\[3mm]&\color{#ff0000}{\geq \int_{0}^{1}\int_{0}^{1}\dd x\,\dd y = 1}
\end{align}

$$
\color{#0000ff}{\int_{0}^{1}\fermi\pars{x}\,\dd x\int_{0}^{1}{1 \over \fermi\pars{x'}}\,\dd x'}
\geq \color{#0000ff}{1}
$$

