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Let $f(x)\in C^{1}(\mathbb{R}),\ f^{'}(x)>f(f(x)),\ \forall x\in\mathbb{R}. $
Prove: $f(f(f(x))))\le 0,\ \forall x\ge0.$
I have just proved $f$ is bonuded. In fact, if $\displaystyle\lim_{x\to +\infty}f(x)=+\infty,\ $ when $x$ is sufficiently big, $$f(x)>1+\epsilon>1,\ f^{'}(x)>f(f(x))>1+\epsilon$$so that when x is more sufficiently big, we have$$f(x)>x,\ f^{'}(x)>f(f(x))>f(x),\ e^{-x}f(x)>c>0,\ f^{'}>ce^{f(x)}$$than$$+\infty>\int_{x_0}^{+\infty}\frac{f^{'}(x)}{e^{cf(x)}}=+\infty$$that's a contradiction!
But then I ran out of ideas.

Piquancy
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    An Idea, I define: $f^{(n)} = \underbrace{f\circ \cdots \circ f}_{n \text{ times}}$. We have $f'(f^{(n)}(x) > f^{(n+2)}(x)$. Suppose $f^{(3)}(x) > 0, x \geq 0$, and compute $df^{(3)}/dx$. By applying the assumption, you obtain $f'(f^{(2)}(x))f'(x) > 0$. You can deduce a couple of things from there. – Pastudent Feb 03 '23 at 14:57
  • Someone else asked the same question here in 2017, however there was no answer there either. Would you mind sharing the source of the question? Just to see if anything pops up anywhere else. – Bruno B Feb 03 '23 at 15:24
  • We might be able to retrieve some of the stuff that is in here (from AoPS)? It deals with what happens when we want equality instead of $>$, but maybe some of the ideas could be reused? – Bruno B Feb 03 '23 at 15:39
  • @Pastudent Thank you! – Piquancy Feb 03 '23 at 15:52
  • @BrunoB I just saw it a few days ago. Others told me that it came from IMC. But I read all the tests and couldn't find this problem in AoPS. – Piquancy Feb 03 '23 at 15:55
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    Found it, it's problem 4 of the first day of IMC 2012 (link here) – Bruno B Feb 03 '23 at 16:01
  • Ok, I may have missed it. – Piquancy Feb 03 '23 at 16:06

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