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$\begingroup$

I tried differentiating it and get a whole mess:

$f'(p)=(u^p+1)^{1/p}\left(\frac{u^p\log{u}}{p(u^p+1)}-\frac{\log{(u^p+1)}}{p^2}\right)$

And I don't know how to prove that this is always negative when $u>0$ and $p>1$.

1 Answers 1

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(I trust your differentiation.) In the parentheses, it is

$\frac{u^p\log u^p - (u^p + 1)\log(u^p + 1)}{p(u^p + 1)}.$

Just look at the numerator.

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    That's nice! Put $p$ in the $\log{}$! Thank you.2012-09-07