For any natural $m \neq n$ show that $|\sqrt[n]{m} - \sqrt[m]{n}| > \frac{1}{mn}$. Here's my try.
The inequality above is equivalent to $$|m^{\frac{1}{n}} - n^{\frac{1}{m}}|> \frac{1}{mn}$$ First, I want to get rid of the absolute value.
Assume without loss of generality that $m>n$. Then $m^m > n^n$. Raising this inequality to the power of $\frac{1}{mn}$ gives me $m^{\frac{1}{n}} > n^{\frac{1}{m}}$. 
Next, I want to arrange everything so that it looks like $n^{\frac1n}-m^{\frac1m}$($m$ goes with $m$, $n$ goes with $n$ $-$ that's easier to consider). For that, I know when the function $f(x) = x^{\frac1x}$ is decreasing. As $f'(x) = (e^{\frac1x \cdot \ln{x}})' = x^{\frac1x - 2}\cdot (1- \ln{x})$, it is clear that $f(x)$ is decreasing on the ray $[e;+\infty )$. So, consider $m$, $n$ $\geq 3$. (Other cases can be checked numerically.)
So, $m>n$, then $m^{\frac{1}{n}} > n^{\frac{1}{n}}$. At the same time $n^{\frac{1}{m}} < m^{\frac{1}{m}}$ $\Rightarrow$ $-n^{\frac{1}{m}} > -m^{\frac{1}{m}}$. Adding up these two inequalities gives me $$m^{\frac{1}{n}} - n^{\frac{1}{m}} > n^{\frac{1}{n}} - m^{\frac{1}{m}} > 0$$
Secondly, I want to utilize the fact that $m$ and $n$ are natural. As $m>n$, then $m-n\geq 1$. Dividing by $mn$ gives me $$\frac{m-n}{mn} = \frac1n - \frac1m \geq \frac1{mn}$$
Now I just have to prove this:
$$n^{\frac{1}{n}} - m^{\frac{1}{m}} > \frac1n - \frac1m$$
Or, equivalently, $$ n^{\frac{1}{n}} - \frac1n > m^{\frac{1}{m}} - \frac1m$$
So, consider the function $g(x) = x^{\frac1x} - \frac1x $. I want to show that it is decreasing, starting at some point. Taking the derivative does nothing good: $g'(x) = \frac1{x^2} \cdot (x^{\frac1x} - \ln{x} \cdot x^{\frac1x} + 1)$, and I am unable to find the zeroes of it. WolframAlpha says that $g(x)$ is, indeed, decreasing from $x\approx 5.677$, and that is good, we can always check the answer for $m$, $n$ $\leq 5$. However, this is not a satisfactory solution.

Suppose that $m>n$, as a result of which $m^m>n^n$. If $m=2$, then $n=1$ and the proof is easy to complete; suppose thus that $m\ge 3$.
Applying Lagrange's mean value theorem to the function $f(x)=x^{1/(mn)}$ on the interval $[n^n,m^m]$, we get
  $$ \sqrt[n]m-\sqrt[m]n = f(m^m)-f(n^n) = \frac1{mn}\,c^{-1+1/(mn)}(m^m-n^n), $$
where $c\in(n^n,m^m)$. Consequently, 
  $$ \sqrt[n]m-\sqrt[m]n > \frac1{mn}\,m^{-m+1/n}\, (m^m-n^n), $$
and to complete the proof it suffices to show that 
  $$ m^m-n^n > m^{m-1/n}. $$
Clearly, this will follow from 
  $$ m^m-n^n>m^{m-1/m}, $$
and then, in view of $m^{1/m}>1+1/m$ (which is very easy to prove), from 
  $$ \left(1+\frac1m\right)(m^m-n^n)>m^m. $$
The last inequality simplifies to 
  $$ m^{m-1} > \left(1+\frac1m\right)n^n, $$
and this is true in view of
  $$ m^{m-1}\ge (n+1)^n \ge \left(1+\frac1n\right)n^n > \left(1+\frac1m\right)n^n. $$
