How to show $\log a\le n(\sqrt[n]{a}-1) \le \sqrt[n]{a}\log a$

Question

Let $ b=\sqrt[n]{a}$.

How to show: $\log a\le n(\sqrt[n]{a}-1) \le \sqrt[n]{a}\log a$?

Thank you ;)

Answer 1

Noting $a = b^n$, this is equivalent to $$n \log b \leq n(b - 1) \leq n b \log b,$$ i.e. just $$\log b \leq b -1 \leq b \log b.$$ To get this, employ the mean value theorem separately for $b \geq 1$ and $0 < b < 1$ (I assume the condition $b>0$ as $\log$ 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 $x \mapsto \log x$ gives us that there exists $c \in [1,b]$ such that $$\log b - \log 1 = \frac{1}{c}(b-1),$$ then noting $\log 1 = 0$, and using the facts $c \in [1,b]$ and $b-1$ are positive gives us $$\frac{1}{b}(b-1) \leq \frac{1}{c}(b-1) \leq b-1$$ 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 $c\in [b,1]$ and $b-1$ is negative instead.