Assuming the Fermat Theorem, show that there is no natural number $x$, $y$, and $z$ and $n\geq3$ such that $$\frac{1}{x^n} + \frac{1}{y^n} = \frac{1}{z^n}. $$
So far I think proof by contradiction may be the best route, but I cannot find any place where I can start.
On
The only property of Fermat's Last Theorem used in the problem is that
FLT with exponent $n$ is a homogeneous equation; if $(a,b,c)$ is a solution then $(ka,kb,kc)$ is a solution, for any $k$.
This makes the existence of solutions in rational numbers equivalent to the existence of integer solutions (because an integer solution is a rational solution, and clearing denominators in a rational solution gives an integer solution).
Suppose to the contrary that $a,b,c$ are natural numbers such that $\frac{1}{a^n}+\frac{1}{b^n}=\frac{1}{c^n}$. Multiply through by $a^nb^nc^n$. We get $(bc)^n +(ac)^n=(ab)^n$. Thus $x=bc$, $y=ac$, $z=ab$ would be a natural number solution of the equation $x^n+y^n=z^n$. This contradicts the fact (Fermat's Last Theorem, proved by Wiles) that for $n\ge 3$, the equation has no solution in natural numbers.