# 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 $$0 for each $$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!

• Do you know the Cauchy-Schwarz inequality? Jun 11 '21 at 16:33
• Jun 11 '21 at 16:34

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