Inequality proof: If $ m>3 $ then $ \sqrt [m] {m} < \sqrt [3] {3} $

Question

How would you go about proving the following using inequalities:

If $$ m>3 $$ Where $$ m \in \mathbb{R} $$ then $$ \sqrt [m] {m} < \sqrt [3] {3} $$

I've been trying to prove this for the past few days and have made little progress. I've tried taking the mth root or the cube root of both sides of the original inequality but this does not seem to help: $$ m>3 $$ $$ \sqrt [m] {m} > \sqrt [m] {3} $$ I'm not sure where to go from here.

I've also tried taking logarithms at both sides: $$ \ln(m) > \ln(3) $$ $$ \frac {1}{m} \ln(m) > \frac {1}{m} \ln(3) $$ and $$ \frac {1}{m} \ln(3) < \frac {1}{3} \ln(3)$$ which gives $$ \sqrt [m] {3} < \sqrt [m] {m}$$ $$\sqrt [m] {3} < \sqrt [3] {3}$$but I can't figure out a way to manipulate this further to prove $$ \sqrt [m] {m} < \sqrt [3] {3} $$

I feel like I'm missing something obvious, but the inequities just seem awkward to work with. I think you could prove it using the function $f(x) = x ^ {\frac {1}{x}}$ and proving its derivative is negative for $x>3$, but I was wondering if there was a way to prove this using inequalities, or is it not possible to prove it this way.

Any help would be much appreciated.

Answer 1

Consider the function $f: \mathbf{R}^+ \to \mathbf{R}^+$ defined by $x\mapsto x^{1/x}$. Then $$ f^\prime(x)=\frac{\mathrm{d}}{\mathrm{d} x}e^{\frac{\ln x}{x}}=f(x)\frac{\frac{1}{x}x-\ln x}{x^2}=\underbrace{\frac{f(x)}{x^2}}_{>0}(1-\ln x). $$ Therefore $f$ is increasing if and only if $x \in (0,e)$.

Answer 2

HINT

Let $f(m) = m^{1/m}$ for $m>3$. Prove that $f(m)$ is a decreasing function. (It may help to note that $f(m)$ is increasing if $g(m) = \ln f(m)$ is increasing...)