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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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    @Byron It doesn't matter who posts an answer as long as it's correct and OP understands it :)2011-09-02

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

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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.2011-09-02