Nelson Jennings

2022-07-16

How to show $\mathrm{log}a\le n\left(\sqrt[n]{a}-1\right)\le \sqrt[n]{a}\mathrm{log}a$
Let $b=\sqrt[n]{a}$
How to show: $\mathrm{log}a\le n\left(\sqrt[n]{a}-1\right)\le \sqrt[n]{a}\mathrm{log}a$?
Thank you ;)

tiltat9h

Noting $a={b}^{n}$, this is equivalent to
$n\mathrm{log}b\le n\left(b-1\right)\le nb\mathrm{log}b,$
i.e. just
To get this, employ the mean value theorem separately for $b\ge 1$ and $0 (I assume the condition $b>0$ as $\mathrm{log}$ isn't defined for non-positive numbers). Notice first it does work for $b=1$, as all 3 quantities are $0$. If $b>1$, applying the mean value theorem to $x↦\mathrm{log}x$ gives us that there exists $c\in \left[1,b\right]$ such that
$\mathrm{log}b-\mathrm{log}1=\frac{1}{c}\left(b-1\right),$
then noting $\mathrm{log}1=0$, and using the facts $c\in \left[1,b\right]$ and $b-1$ are positive gives us
$\frac{1}{b}\left(b-1\right)\le \frac{1}{c}\left(b-1\right)\le b-1$
gives us the result we want after a little bit of re-arranging. For $b<1$, the working is pretty much the same, except we get the last inequality from the fact $c\in \left[b,1\right]$ and $b-1$ is negative instead.

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