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. 
A: We have:
$\sqrt[n]{m} > \sqrt[3]{3}$ and $\sqrt[m]{n} > \sqrt[3]{3}$.  
Expressed using fractional powers, these are precisely:
${m^{1/n}} > {3^{1/3}}$ and ${n^{1/m}} > {3^{1/3}}$.  
Taking nth and mth powers respectively, we get:
${m > 3^{n/3}}$ and ${n > 3^{m/3}}$.  
Substitute for m from the first inequality in the second:
${n > 3^{{3^{n/3}}/3}}$.  
Simplifying the exponent:
${n > 3^{3^{(n-3)/3}}}$.  
Could this be true for all n?  Take, for example, n=303, giving:
${303 > 3^{3^{100}}}$
which is patently false.  So we have our contradiction.
[Edit]  Of course, the above merely shows that for some m, there exist values of n for which the two roots n^(1/m) and m^(1/n) can't both exceed 3^(1/3).  However, I think we should read the original question as asking us to show that for every pair of integers m, n both > 1, with each integer being chosen independently of the other, those two roots can't both exceed 3^(1/3).  The reasoning I gave showed much less than that!  Although I've since tried other approaches, I now believe that the only other Answer to date, by user S. Y, uses probably the best and simplest approach.
