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

Use Fubbini's theorem on double integral $\iint_{(0,1)\times (0,1)}g(u)\cdot f(v) du dv.$ –  אליהו צלע Jan 5 '13 at 21:51
$$\int_0^1 \sqrt{f} \sqrt{g} \mathrm dx \leq \sqrt{\int_0^1f \mathrm dx \int_0^1 g \mathrm dx}$$