$x^n+y^n=z^n$ where $x,y,z$ are real numbers Problem:
 If $$ x^n+y^n=z^n$$ where $x,y $ and $ z $ are real numbers $\gt 0$ and n is any integer $ \neq 0$ then  $$ \frac {x}{n} + y >z$$ assuming $x \lt y $
Background: While studying Fermat's theorem I saw that the following are the characteristics of $x,y$ and $z$ $$ x+y \gt z\gt y \gt x$$ But while studying further, I also saw the above to be true.
Request: I could prove this when $n \gt 0$ but couldn't do the same for $n \lt 0$. If anyone could prove or disprove both the cases or show me any site where this has been discussed. A detailed proof would be appreciated. I shall share my proof shortly for a review for the case $n \gt0$
 A: Here is a proof for $n \geq 1$, unfortunately for $n<0$ we get a weaker ienquality.
For all $n \notin(0,1)$ we have
$$(y+\frac{x}{n})^n = y^n(1+\frac{x}{ny})^n > y^n(1+\frac{x}{y})$$
with the last inequality following from Generalized Bernolli
For $n \geq 1$ you also have:
$$ y^n+xy^{n-1}> y^n+x^n =z^n\,,$$
and by combining these two you get the desired result.
Maybe someone can fix it in general.

Added: Inspired by only's answer.
If $n=-1$, your equation is
$$\frac{1}{x}+\frac{1}{y}=\frac{1}{z} \Rightarrow xz+yz+=xy \Rightarrow z= \frac{xy}{x+y}$$
You need $y-x >z$, or equivalently
$$y-x > \frac{xy}{y+x} \Leftrightarrow y^2 > x^2+xy$$
Note that $y=3, x=2$ is a counterexample to this equation, thus $y=3, x=2, z=\frac{6}{5}, n=-1$ is  a counterexample to your claim.
In general, if $n=-m<0 $ you want to prove that
$$\frac{1}{x^m}+\frac{1}{y^m} =\frac{1}{z^m} \mbox{implies} \, y-\frac{x}{m} >z$$
but since
$$\frac{z^m}{y^m}+\frac{z^m}{x^m}=1\,,$$
if you set $\frac{z^m}{y^m} =\frac{1}{2}+\alpha$ and $\frac{z^m}{x^m} =\frac{1}{2}-\alpha$, the desired inequality becomes
$$\frac{1}{\sqrt[m]{\frac{1}{2}+\alpha}}-\frac{1}{m\sqrt[m]{\frac{1}{2}-\alpha}} >1$$
Since this is increasing in $\alpha$, is easy to see that this is true for all $0 < \alpha < \frac{1}{2}$ if and only if
$$\frac{m-1}{m\sqrt[m]{\frac{1}{2}}}  =\lim_{\alpha \to 0^+} \frac{1}{\sqrt[m]{\frac{1}{2}+\alpha}}-\frac{1}{m\sqrt[m]{\frac{1}{2}-\alpha}} \geq 1$$
or equivalently
$$2 \geq (\frac{m}{m-1})^m $$.
But this is not possible, since the RHS is a sequence which is decreasing to $e$. This shows that for each $m$ there exists some $\alpha$ which makes the inequality fail....
A: $n = -1, x = y = 2, z = 1$ is a solution (I know you require $x < y$; just decrease x by a bit and increase y by a bit.)
Then, we can make $y-x$ arbitrarily close to 0, while z = 1, so this disproves your assertion for $n = -1.$ 
A: 1) The estimates pertaining to fermat's equation have been extensively studied. One may see  this paper
http://www.fq.math.ca/Scanned/29-1/bialek.pdf
2) Your problem is a theorem of Grunert, and it admits a simple proof.
Hope these helpful.
Goodluck!
