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Let $f:(0,\infty) \rightarrow \mathbb{R}$ be continuous. I need to show that

$$\left(\int_1^ef(x)dx \right)^2 \leq \int_1^e xf(x)^2dx$$ I have been trying to use C-S to prove this but with no luck.

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what did you try? There's only one way to apply Cauchy Schwarz here and it would work. Hint: What should you multiply to $xf(x)^2$ so that the square root of product is $f(x)$? –  Soarer Sep 1 '11 at 21:54
    
@Soarer I used the fact $\sqrt{x} \geq 1$ and got as far as I mention in response to DJC –  user9352 Sep 1 '11 at 21:55
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@Soarer I posted a hint as an answer before I saw your comment. Now I deleted it. –  Srivatsan Sep 1 '11 at 21:58
    
@user9352, you may try to answer the hint I mentioned in the last comment. –  Soarer Sep 1 '11 at 21:58
    
thanks I get it now –  user9352 Sep 1 '11 at 22:03

1 Answer 1

up vote 11 down vote accepted

$$\left(\int_1^ef(x)dx \right)^2 = \left(\int_1^e{1\over\sqrt{x}}\cdot \sqrt{x}f(x)dx \right)^2 \leq \int_1^e{1\over x}\,dx\cdot\int_1^e xf(x)^2dx=\int_1^e xf(x)^2dx$$

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I would just like to mention for completeness that the inequality is due to Cauchy-Schwarz and the equality on the right comes from integrating 1/x. –  user9352 Sep 2 '11 at 14:35

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