Prove $\sqrt[n]{m}\leq\sqrt[3]{3}\lor\sqrt[m]{n}\leq\sqrt[3]{3}$ for $n,m\in\mathbb{N}>1$.

Solution 1:

You want to show that $$ \min\left({\sqrt[m]{n}, \sqrt[n]{m}}\right) \le \sqrt[3]{3}. $$ Raising both sides to the $(3mn)$-th power, we see that this is equivalent to $$ \min\left(m^{3m}, n^{3n}\right)\le 3^{mn}. $$ Assume with no loss of generality that $m \le n$; then $\min\left(m^{3m},n^{3n}\right)=m^{3m}$ and $3^{mn}\ge 3^{m^2}$. Because $ m^3 \le 3^m $ for all $m\ge 1$ (*), we have $$ m^{3m} =\left[m^3\right]^{m}\le \left[3^{m}\right]^{m} = 3^{m^2} \le 3^{mn}$$ and we're done.

(*) The proof that $m^3 \le 3^m$ for all $m \ge 1$ is by induction: it holds for $m=1,2,3$ by direct inspection, and in going from $m$ to $m+1$ for $m\ge 3$, the right-hand side is multiplied by $3$, while the left-hand side is multiplied by $$\frac{(m+1)^3}{m^3}=1+\frac{3}{m}+\frac{3}{m^2}+\frac{1}{m^3} \le 1+\frac{3}{3}+\frac{3}{3^2}+\frac{1}{3^3}<3.$$

Solution 2:

Strictly speaking, it fails for $m=n=3$ unless you make your $\lt$ into $\le$. If you prove it for $n=m$, you can argue that for $n \gt m, \sqrt[n]m \lt \sqrt[m]m$, so you only need to worry about equality. Using calculus, you can show that $\sqrt[x]x$ is decreasing for $x \gt 3$ and you are there.