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I want to show $[(f^k)^{(n)}]^2 \geq (f^{k-1})^{(n)} (f^{k+1})^{(n)}$, where $f$ satisfies $f^{(n)} \geq 0$ for all integer $n$ and $f^k$ denotes the $k$-th power of $f$.

I believe it's right but I can't see it clearly.

Thanks.

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  • $\begingroup$ This is not true for at least $n\ge 3$. I'm trying to cook up an example. $\endgroup$ Feb 27, 2014 at 2:30

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Take $n=3, k=2$, then your inequality will be equivalent to show the following:

$36{f'}^2{f''}^2+24ff'f''f'''+4f^2{f'''}^2 \ge 6{f'}^3f'''+18ff'f''f'''+3f^2{f'''}^2$,

so equivalent to $36{f'}^2{f''}^2+6ff'f''f'''+f^2{f'''}^2 \ge 6{f'}^3f'''$.

Now pick some function $f(x)$, such that $f'$ is big enough compared to $f,f''$ and $f'''$ at least at some point and then your inequality will not hold at that point.

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  • $\begingroup$ Specifically, you can take $f(x)=x+x^3$ and then the inequality fails at $x=0$. $\endgroup$ Feb 27, 2014 at 7:53

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