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$f$ is a bounded and continuous function with $f:D \to (0,\infty)$ on the Jordan-measurable set. Prove that inequality $$\int_Df(x)dx\int_D{1\over f(x)}dx\ge(vol(D))^2$$

Can someone please help me with this? I really am stuck on this Problem.

Any help would be greatly appreciated.

suugii19
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1 Answers1

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From the Cauchy Schwarz Inequality, we see $$\text{vol}(D) = \int_D 1\, dx = \int_D \sqrt{f(x)} \frac{1}{\sqrt{f(x)}} dx \le \left( \int_D f(x) dx \right)^{1/2} \left(\int_D \frac 1 {f(x)} dx\right)^{1/2}$$ which is exactly what you needed to prove.

User8128
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