Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
@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
up vote 11 down vote accepted

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.