Problem: Let $x,y,z > 0$ satisfy $xyz\geq 1$. Prove that
\[ \frac { x^5-x^2 }{x^5+y^2+z^2}
  + \frac {y^5-y^2}{x^2+y^5+z^2}
  + \frac {z^5-z^2}{x^2+y^2+z^5} \geq 0. \]

Solution: Negating both sides and adding $3$ eliminates the minus signs:
\[ \sum_{\text{cyc}} \frac{1}{x^5+y^2+z^2}
  \le \frac{3}{x^2+y^2+z^2}. \]
Thus we only need to consider the case $xyz = 1$.

Direct expansion and Muirhead works now!
As advertised, once we show it suffices to analyze if $xyz=1$
the inequality becomes more economically written as
\[ S = \sum_{\text{cyc}} x^2(x^2-yz)(y^4+x^3z+xz^3)(z^4+x^3y+xy^3)
  \overset{?}{\ge} 0. \]
So, clearing all the denominators gives
\begin{align*}
  S &= \sum_{\text{cyc}} x^2(x^2-yz)
  \left[ y^4z^4 + x^3y^5 + xy^7 + x^3z^5 + x^6yz + x^4y^3z
    + xz^7 + x^4yz^3 + x^2y^3z^3 \right] \\
  &= \sum_{\text{cyc}}
  \left[ x^4y^4z^4 + x^7y^5 + x^5y^7 + x^7z^5 + x^{10}yz + x^8y^3z
    + x^5z^7 + x^8yz^3 + x^6y^3z^3 \right] \\
  &- \sum_{\text{cyc}}
  \left[ x^2y^5z^5 + x^5y^6z + x^3y^8z + x^5yz^6 + x^8y^2z^2 + x^6y^4z^2
    + x^3yz^8 + x^6y^2z^4 + x^4y^4z^4 \right] \\
  &= \sum_{\text{cyc}}
  \left[ x^7y^5 + x^5y^7 + x^7z^5 + x^{10}yz
    + x^5z^7 + x^6y^3z^3 \right] \\
  &- \sum_{\text{cyc}}
  \left[ x^2y^5z^5 + x^5y^6z + x^5yz^6 + x^8y^2z^2 + x^6y^4z^2
    + x^6y^2z^4 \right]
\end{align*}
In other words we need to show
\[
  \sum_{\text{sym}} \left( 2x^7y^5
  +\half x^{10}yz + \half x^6y^3z^3 \right)
  \ge
  \sum_{\text{sym}} \left( \half x^8 y^2 z^2
  + \half x^5 y^5 z^2 + x^6 y^4 z^2 + x^6 y^5 z \right).
\]
which follows by summing
\begin{align*}
  \sum_{\text{sym}} \frac{x^{10} yz + x^6 y^3 z^3}{2}
    &\ge \sum_{\text{sym}} x^8 y^2 z^2 \\
  \half \sum_{\text{sym}} x^8 y^2 z^2
    &\ge \half \sum_{\text{sym}} x^6 y^4 z^2 \\
  \half \sum_{\text{sym}} x^7y^5
    &\ge \half \sum_{\text{sym}} x^5 y^5 z^2 \\
  \half \sum_{\text{sym}} x^7y^5
    &\ge \half \sum_{\text{sym}} x^6 y^4 z^2 \\
  \sum_{\text{sym}} x^7y^5
    &\ge \sum_{\text{sym}} x^6 y^5 z.
\end{align*}
The first line here comes from AM-GM,
the rest come from Muirhead.
\begin{remark*}
More elegant approach is to use Cauchy in the form
\[ \frac{1}{x^5+y^2+z^2} \le \frac{x\inv+y^2+z^2}{(x^2+y^2+z^2)^2}. \]
\end{remark*}