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Let $f:[0,1]\to\Bbb R$ be a measurable function satisfying $0<f(x)<\infty$ for each $x\in [0,1]$. Show that $\int_0^1 f(x)dx\int_0^1\frac{1}{f(x)}\geq 1$.


$Attempt$. Since $f>0$ there exists a sequence $\{\phi_n\}_{n=1}^\infty$ of simple functions on $[0,1]$ which converges pointwise on $[0,1]$ to $f$ and such that $0<\phi_1\leq\phi_2\leq \dots$ We observe that if the statement is true for any simple function $\phi:[0,1]\to\Bbb R$ with $\varphi>0$ then $1\leq \int_0^1\phi_n(x)dx\int_0^1\frac{1}{\phi_n(x)}dx$ implies that $$1\leq \lim_{n\to\infty}\int_0^1 \phi_n(x)dx\lim_{n\to\infty}\int_0^1 \frac{1}{\phi_n(x)}dx\stackrel{?}{=}\int_0^1 f(x)dx\int_0^1\frac{1}{f(x)}dx$$ by Monotone Convergence Theorem. But, I couldn't prove it for any positive simple function. Any help would be appreciated. Thanks!

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Since $f$ takes values greater than 0 we have $f(x) (1/f(x))=1$ for all $x. $ So taking the square root we also have $(f(x)) ^{1/2}(1/f(x))^{1/2} =1 $.

Then by Cauchy Schwarz $1=\int (f(x)) ^{1/2}(1/f(x))^{1/2}\ dx \leq (\int f(x) \ dx) ^{1/2} (\int (1/f(x)) \ dx) ^{1/2}. $

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There is a neat answer using only the elementary inequality $x+\frac{1}{x}\geq 2$ for $x>0$, which immediately follows from $\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2\geq 0$.

Namely, using Fubini's theorem, we can write $$ \begin{split} 2\left(\int_0^1 f(x)\mathrm{d}x\right)\left(\int_0^1 \frac{1}{f(x)}\mathrm{d}x\right)&=\left(\int_0^1 f(x)\mathrm{d}x\right)\left(\int_0^1 \frac{1}{f(y)}\mathrm{d}y\right)+\left(\int_0^1 f(y)\mathrm{d}y\right)\left(\int_0^1 \frac{1}{f(x)}\mathrm{d}x\right) \\ &=\int_0^1\int_0^1 \frac{f(x)}{f(y)}\mathrm{d}x\mathrm{d}y+\int_0^1\int_0^1 \frac{f(y)}{f(x)}\mathrm{d}x\mathrm{d}y \\ &=\int_0^1\int_0^1 \left(\frac{f(x)}{f(y)}+\frac{f(y)}{f(x)}\right)\mathrm{d}x\mathrm{d}y \\ &\geq \int_0^1\int_0^1 2\,\mathrm{d}x\mathrm{d}y=2 \end{split} $$ and hence we are done.

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