# $f$ decreasing monotonically decreasing , Prove: for every $\alpha \in (0,1)$ :$\int_{0}^{\alpha} f(x)dx \geq \alpha\int_{0}^{1}f(x)dx$

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.

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

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

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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.

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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.

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