Show that $\int_0^1 f(x)dx\int_0^1\frac{1}{f(x)}\geq 1$ 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!

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}. $
