Q:

Prove that at least one of the expressions does not exceed $\sqrt[3]{3}$

Let $m,n > 1$ be positive integers. Prove that at least one of the numbers $$\sqrt[n]{m} \ , \ \sqrt[m]{n}$$ does not exceed $\sqrt[3]{3}$.

I thought about doing a proof by contradiction. That is, assume this is not the case and so both exceed $\sqrt[3]{3}$. Then, $\sqrt[n]{m} > \sqrt[3]{3}$ and $\sqrt[m]{n} > \sqrt[3]{3}$. How do we continue?

A:

Without loss of generality, assume that $m\ge n$. Then $$\sqrt[m]{n} \le \sqrt[n]{n}$$
We just want to show that $$\sqrt[n]{n} \le \sqrt[3]{3}, $$
which is equivalent to 
     $$n^3 \le 3^n.$$
This is not hard by,   for example, proof by induction. 
When $n=2$ or $3$, this is true, and when $n > 3$, $$n^3 = (n-1)^3 + 3(n-1)^2+ 3(n-1) + 1  \le (n-1)^3 + (n-1)^3 + 3(n-1) + 1 = 2(n-1)^3 + 3n -2 < 2(n-1)^3 + (n-1)^3 = 3(n-1)^3 \le 3*3^{n-1} = 3^n$$
If you want to use calculus, it is to prove that $$x^{\frac{1}{x}}$$ is decreasing when $x\ge e$, which is also not hard. 

