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Suppose $m(E)=1$ and $f,g \geq 0$ with $fg \geq 1$. Show that $\displaystyle \int_E f dm \int_E g dm \geq 1$

Can someone please point me to the right direction?

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I see that you have not accepted answer to any of your previous 5 questions. If you think a particular answer has helped you to solve a problem, kindly accept it by checking the tick mark to the left of the question. –  user17762 Feb 12 '11 at 19:12

2 Answers 2

up vote 2 down vote accepted

Hint : Use Cauchy-Scharwz inequality.

Hint 2 : If $f,g \geq 0, fg \geq 1$ then $\sqrt{f}\sqrt{g} \geq 1$ also.

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thanks i think i got it –  jack Feb 12 '11 at 0:37
    
Good. If you want to be sure : Applying integral CS to $\sqrt{f}$ and $\sqrt{g}$, you have $\int_E f \int_E g \geq \left(\int_E \sqrt{f}\sqrt{g}\right)^2 \geq \int_E dt = 1$. –  Sam Feb 12 '11 at 0:41

I assume that $f$ and $g$ are integrable. Outside a null-set we have $g \geq \frac{1}{f} \gt 0$, in particular $\frac{1}{f}$ is integrable. Now apply Jensen's inequality $\varphi\left(\int f\right) \leq \int \varphi \circ f$ for the convex function $\varphi(x) = \frac{1}{x}$ (here we use $f \geq 0$ again), so $\frac{1}{\int f} \leq \int \frac{1}{f} \leq \int g$.

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