Prove that $\int_0^1f(x)dx$$\int_0^1g(x)dx\ge1$.

Question

Suppose $f(x)$ and $g(x)$ are positive measurable functions defined on $(0,1)$, satisfying $f(x)g(x)\ge1$ for any $x\in(0,1)$. Prove that $\int_0^1f(x)dx$$\int_0^1g(x)dx\ge1$.

Totally no idea about this problem. Any hints? Is there any theorem I'm supposed to apply?

Answer 1

Apply Cauchy Schwarz.
$$\int_0^1 \sqrt{f} \sqrt{g} \mathrm dx \leq \sqrt{\int_0^1f \mathrm dx \int_0^1 g \mathrm dx}$$