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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?

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Use Fubbini's theorem on double integral $\iint_{(0,1)\times (0,1)}g(u)\cdot f(v) du dv.$ –  Elias Jan 5 '13 at 21:51

1 Answer 1

up vote 7 down vote accepted

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

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