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Let $f(x)$ be a function from $C^3( \mathbb{R} )$ satisfying the following conditions for all $x \in \mathbb{R}$: $$f(x)<0, f'(x)>0, f''(x)<0, f'''(x)>0, 27f'''(x)+f(x) \leq 0.$$ Prove that $3f'(x)+2f(x)<0$.

Source. This problem is from mathematical competition in Moscow that was held in 2023.

My attempt.

  • I tried to make a substitution $f(x)=g(\frac{x}{3})$ $\big($then $27f'''(x)=g'''(\frac{x}{3}) \big)$.
  • I also tried to use somehow that $0>f'f \Rightarrow 0>f^2(x)-f^2(0)$ and the same for $f'f''<0$ and $f''f'''<0$.
  • I denoted $$h(x)=3f'(x)+2f(x) \tag{*}$$ and reduced the problem to proving $h(x)<0$, where $f$ may be found as $$f(x)=e^{-\frac{2x}{3}}\left(f(0)+\frac{1}{3} \int_{0}^{x}e^{\frac{2t}{3}}h(t)dt \right).$$ Moreover from $(*)$ we have $f=-\frac{3}{2}f'+\frac{1}{2}h(x)<0$ and thus $h(x)<3f'(x)$ $\big($same for $f'=-\frac{3}{2}f''+\frac{1}{2}h'(x)>0$ and further for $f''...\big)$, but that wasn't successfully.

I don't know what an expression/identity/substitution I should guess to solve the problem. Help me please

River Li
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pioo
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    Just in case it might help with finding the right path to the solution, where did you find this problem? For example, was it a contest question? – Bruno B Nov 18 '24 at 17:04
  • @BrunoB this problem is from mathematical competition in Moscow that was held in 2023 – pioo Nov 19 '24 at 22:44
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    I forgot which one - but one year of the IMC has a very similar problem. – dezdichado Dec 19 '24 at 14:24
  • @dezdichado I saw all the problems from the IMC from every year and I don't remember anything that might be really useful to solve this problem – pioo Dec 21 '24 at 09:29
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    @pioo hmmm I must have misremembered then - apologies for the mislead. But I definitely have seen the first four of the five requirements on $f$ on a problem somewhere. – dezdichado Dec 21 '24 at 15:54
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    Let $f(x)=-\exp g(x)$ and $h(x)=1/g'(x)$. Then the conditions are equivalent to $$h(x)<0,\quad h'(x)<1,\quad\frac{h(x)^3}{27}\le h(x)h''(x)-2h'(x)^2+3h'(x)-1<0$$ for all $x\in\Bbb R$ and we wish to show that $h(x)<-3/2$. – TheSimpliFire Dec 21 '24 at 16:06

1 Answers1

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As @dezdichado remembered, this is a near-duplicate of question B4 in the 1999 Putnam exam, which is already answered here on MSE or here on the Putnam website.

Here are the details. Put $g(t)=-f(-3t)$. Then $g^{(k)}(t)=-(-3)^kf^{(k)}(-3t)$, and hence $g,g',g''$ and $g'''$ are all nonnegative.

Condition $f(x)+27f'''(x) \leq 0$ is equivalent to $g(t)\geq g'''(t)$.

Condition $3f'(x)+2f(x) \lt 0$ is equivalent to $g'(t)\leq 2g(t)$.

But, by any of the linked answers above, we indeed do have $g'(t) \leq 2g(t)$, qed.

Ewan Delanoy
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