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True or False?

Let $f$ be a differentiable function on $\mathbb{R}$. If $ f \circ f \circ f $ is increasing and $f \circ f$ is decreasing, then $f$ is decreasing.

Here's what I got:

If $f \circ f$ is decreasing then $(f \circ f)' \le 0$

If $f \circ f$ is increasing then $(f \circ f \circ f)' \ge 0$

$$(f \circ f \circ f)' = f'(f \circ f) \cdot (f \circ f)'$$

From this I conclude that

$$f'(f \circ f) \le 0$$

Is this enough for me to prove that $f$ is decreasing?

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    Sounds correct to me. Alternatively, suppose $f$ is increasing. – 2017-02-14
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    I'm mainly having problems with understanding the consequences of the composition on the last part. It doesn't matter if I have f'(x) < 0 or f'(f compose f) < 0? – 2017-02-14
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    @Jan The contrary of $f$ decreasing is not $f$ increasing. – 2017-02-14
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    If $f$ is decreasing, is it possible that $f \circ f$ is decreasing, too? – 2017-02-14
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    @EugenCovaci I think this depens on the terminology (which OP is unclear about as well and is important here). Typically, $f'\geq0$ is called increasing and $f'>0$ strictly increasing. Then the opposite of increasing is NOT decreasing, and you are fully correct. On the other hand, sometimes the convention is $f'\geq0$ non-decreasing and $f'>0$ increasing, so that the opposite of increasing is decreasing. In the current question most likely the first terminology is used judging by the inequalities in the question, but then I don't see how constant $f$ fits into what is to be proven. – 2017-02-14
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    @Jan The definition of increasing/decreasing has nothing to do with the derivative. – 2017-02-14
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    @EugenCovaci I know. But whenever explaining the (increasing and strictly increasing) vs (non-decreasing and increasing) terminology, I need to translate those terms into inequalities and talking in terms of definitions, e.g., $x – 2017-02-14

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From $f'(f \circ f) \le 0$ we get $f'(x) \le 0$ for all $x \in Image(f \circ f)$. Because $Image(f \circ f)$ is not necessarily equals $\mathbb{R}$, you can't conclude $f$ is decreasing on the entire $\mathbb{R}$.

Back to OP question. Actually the statement is false. Suppose $f$ is decreasing. Then $f \circ f$ is increasing, which is false.

It is also false that $f$ is increasing.

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    Ok, so is there another way for me to answer this question? – 2017-02-14
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    @RonH See my updated answer. – 2017-02-14
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    Eugen, I still don't fully understand. Is it a rule that a decreasing function composite of itself is increasing? If it is, why would they give me information about the 3rd composite if it's not needed/ – 2017-02-14
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    @RonH It can be easily proved that the composition of two decreasing functions is increasing. – 2017-02-14
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    So you're basically saying, knowing that, it makes the 3rd composite information obselete? – 2017-02-14
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    @RonH Yes, it doesn't matter. – 2017-02-14