Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'd love your help with proving that for $f:[0,1] \to \mathbb{R}$ monotonically decreasing function ,for every $\alpha \in (0,1)$ :$\int_{0}^{\alpha} f(x)dx \geq \alpha\int_{0}^{1}f(x)dx$. I tried couple of inequalities but I didn't conclude what I should.

Thanks a lot.

share|improve this question
add comment

3 Answers

up vote 6 down vote accepted

$\int_0^{\alpha} f(x) dx = \alpha \int_0^1 f(\alpha t) d t \geq \alpha \int_0^1 f(t) d t$

First equality is integration by substitution with $x=\alpha t$, then inequality holds since $\alpha t \leq t$ and $f$ is decreasing.

share|improve this answer
add comment

You can also note that: $\alpha \int_\alpha^1{f(t)dt}\leq\alpha(1-\alpha)f(\alpha)\leq (1-\alpha)\int_0^\alpha{f(t)dt}$

Rearranging the inequality follows. I think you are able to figure out why the inequalities above holds.

share|improve this answer
add comment

Divide both sides by $\alpha$. The inequality becomes $${1\over \alpha} \int_0^\alpha f(x)\, dx \ge \int_0^1 f(x)\,dx. $$

Define
$$A(\alpha) = {1\over \alpha} \int_0^\alpha f(x)\, dx\qquad \alpha\in[0,1].$$

Then is the average value of $f$ on $[0,\alpha]$. Since $f$ is decreasing, this average must decrease as a function of $\alpha$, so $A(\alpha) \ge A(1)$ for $\alpha\in[0,1]$. The inequality follows immediately.

share|improve this answer
add comment

Your Answer

 
discard

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.