For integers $a < b < c$, is there a general way to sort $a^{1/a}, b^{1/b}, c^{1/c}$?

Question

I'm working through the Math GRE Practice Book and I was struck by problem #6, which asks to sort $\{ 2^{1/2}, 3^{1/3}, 6^{1/6} \}$ in ascending order. I solved it like this:

$$ 2^{1/2} \quad \gtrless \quad 3^{1/3} \quad \gtrless \quad 6^{1/6} $$ $$ 2^{1/2} \quad \gtrless \quad 3^{1/3} \quad \gtrless \quad 2^{1/6} \cdot 3^{1/6} $$ $$ 2^{3/6} \quad \gtrless \quad 3^{2/6} \quad \gtrless \quad 2^{1/6} \cdot 3^{1/6} $$ $$ 2^{2/6} \quad \gtrless \quad \frac{3^{2/6}}{2^{1/6}} \quad \gtrless \quad 3^{1/6} $$ $$ \frac{2^{2/6}}{3^{1/6}} \quad \gtrless \quad \frac{3^{1/6}}{2^{1/6}} \quad \gtrless \quad 1 $$ $$ \Bigl(\frac{4}{3}\Bigr)^{1/6} \quad \gtrless \quad \Bigl(\frac{3}{2}\Bigr)^{1/6} \quad \gtrless \quad 1^{1/6} $$ $$ \frac{4}{3} \quad \gtrless \quad \frac{3}{2} \quad \gtrless \quad 1 $$

and since $ 1 < \frac{4}{3} < \frac{3}{2} $, that tells us that $ 2^{1/2} < 6^{1/6} < 3^{1/3} $.

So, with that problem as inspiration, I'm wondering if there is a known way to generalize this. Like, given three integers $a < b < c$, is there a general way to sort $a^{1/a}, b^{1/b}, c^{1/c}$? In particular, the approach I took depends on 6 having 2 and 3 as factors, so I don't think I'd be able to take the same approach if $a, b, \text{and } c$ are relatively prime.

Answer 1

Let $f : x \mapsto x^{1/x}$. One has $f(x)= \exp \left(\dfrac{\ln(x)}{x} \right)$.

The function $x \mapsto \dfrac{\ln(x)}{x}$ is increasing over $(0,e]$ and decreasing over $[e,+\infty)$.

Therefore, one has directly $f(3) > f(4) > f(5) > f(6) > \ ...$ and so on.

Moreover, by direct computation, $f(4)=f(2)$, and $f(1)=1$.

So finally, one has the following order : $$\boxed{3^{1/3} > 2^{1/2} = 4^{1/4} > 5^{1/5} > 6^{1/6} > \cdots > 1^{1/1}}$$