Prove that $\forall \, a,b \in \mathbb{N}- \{0,1\}\,\, \wedge \,\,a<b \,\, ; \,\, a^{1/a} > b^{1/b}$

Question

Prove that $\forall \, a,b \in \mathbb{N}- \{0,1\}\,\, \wedge \,\,a<b \,\, ; $

$$\,\, a^{1/a} > b^{1/b}$$

I need some tip to start it.

Thank you.

Answer 1

You actually need to assume that the numbers are larger than $2$ for this to be true, since $2^{1/2} < 3^{1/3}$. Regardless, here is a way to proceed with calculus: the function $$f(x) = x^{1/x}$$ has $$\ln f(x) = \frac{\ln x}{x}$$

Taking a derivative gives

$$\frac{f'(x)}{x^{1/x}} = \frac{1 - \ln x}{x^2}$$

Hence $f'$ is negative for $x > e \approx 2.7$, and the result follows.