# Another inequality with definite integrals

Let be $f: \mathbb{R} \rightarrow \mathbb{R}$ a continuous, monotone function. Then, if $a>0$ i must prove that the following inequality holds: $$\int_{-a}^{a}xf(f(x)) \geq0$$

I wonder if there is a simple proof for it.

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We have \begin{align} \int_{-a}^axf(f(x))dx&=\int_0^axf(f(x))dx+\int_{-a}^0xf(f(x))dx\\ &=\int_0^axf(f(x))dx+\int_{a}^0(-s)f(f(-s))(-ds)\\ &=\int_0^as\left(f(f(s))-f(f(-s))\right)ds \end{align} and we deal with two cases:

• $f$ is non-decreasing. Since $-s\leq 0\leq s$, we have $f(-s)\leq f(s)$ and $f(f(-s))\leq f(f(s))$.
• $f$ is non-increasing. We have $f(-s) \geq f(s)$ and $f(f(-s))\leq f(f(s))$.

(in fact $f\circ f$ is non-decreasing in any case)

We can see we have equality if and only if $f\circ f$ is even.

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nice solution. Thanks. – user 1618033 Jun 3 '12 at 20:38
is it correct $f(f(-s))\leq f(s)$ ? I suppose you meant $f(f(-s))\leq f(f(s))$ – user 1618033 Jun 3 '12 at 20:40
@Chris : i think thats typo, should have been $ff(s)$ – Theorem Jun 3 '12 at 20:43
Right, fixed now. – Davide Giraudo Jun 3 '12 at 20:46
@Davide Giraudo: in the last row we have that $f\circ f$ must be odd or even for getting equality? – user 1618033 Jun 3 '12 at 20:54