Prove $\log(x) < n(x)^{1/n}$, for all positive integer values of $n$, and $x > 0$

Question

Given that $$lg(u) < u$$ is always true, how do we use that to prove that $$lg(x) < n(x)^\frac 1n$$

These are the steps that I have taken so far:

$$1: lg(x) < n(x)^\frac 1n$$

$$2: \frac {lg(x)}n < x^\frac 1n$$

$$3: lg(\frac {lg(x)}n) < lg(x)^\frac 1n$$

$$4: lg(\frac {lg(x)}n) < \frac 1n lg(x)$$

$$5: \frac {lg(x)}n < x$$

$$6: lg(x) < nx$$

In which case, since $$lg(u) < u$$ is always true, it thus proves that $$lg(x) < nx$$

I know that I'm probably doing something wrong, especially moving between steps 4 and 5, because the $$\frac 1n$$ shouldn't just disappear like that, but aside from doing this I'm totally stumped.

Any help in understanding this would be greatly appreciated! Thanks, guys!

Answer 1

Step 4 holds because it is $\log u < u$ with $u={\log x\over n}$, and so the proof ends there, by noticing that steps (1)-(4) are actually reversible.

Answer 2

The step from $(4)$ to $(5)$ seems to be the problem. Here is a different approach.

For $u\gt0$, we are given $$ \log(u)\lt u\tag{1} $$ For $n\gt0$, substitute $u\mapsto u^{1/n}$ to get $$ \frac1n\log(u)\lt u^{1/n}\tag{2} $$ Multiply by $n$ to get $$ \log(u)\lt nu^{1/n}\tag{3} $$ The step from $(2)$ to $(3)$ is valid if both sides are greater than $0$. However, $(3)$ is trivially true if $\log(u)\le0$.

Answer 3

Since you start from: $$ \log u < u $$ and the function $x \mapsto x^{1/n}$ is non-negative, set $u = x^{1/n}$. Then: $$ \log x^{1/n} < x^{1/n} $$ and given that $\log x^{1/n} = \frac{1}{n} \log x$, the next step is: $$ \frac{1}{n} \log x < x^{1/n} $$ which gives you the result, since $n$ is positive.

Indeed, there is a tigher inequality: $$ \log x \leq n\left( x^{1/n} -1\right) \leq x^{1/n} \log x $$ with equality attained at $x=1$, as illustrated below for $n=2$:

Answer 4

This way you can prove the inequality :

$$f(x)\ :=\ lg(x)-nx^{\frac{1}{n}}$$

$$f '(x)=\frac{1}{x}-x^{\frac{1}{n}-1}=x^{-1}(1-x^{\frac{1}{n}})$$

The only root of $f '(x)$ is $x=1$. Because of $lim_{x\rightarrow 0} f(x)=-\infty$ and $lim_{x\rightarrow \infty} f(x)=-\infty$, $-n$ is the global maximum of $f(x)$. Hence, $f(x)<0$ for all $x>0$.