Set $p=x^{3/n}$, $q=y^{3/n}$ and $r=z^{3/n}$ with $x,y,z>0$. Then $pqr=1\iff xyz=1$; our goal is to prove \[ \sum \frac{1}{x^3+y^3+1}\le 1. \]Using the fact $x^3+y^3\ge x^2y+xy^2$ (equivalent to $(x-y)^2(x+y)\ge 0$) we get that \[ \sum \frac{1}{x^3+y^3+1}\le \sum\frac{1}{x^2y+xy^2+xyz} = \sum \frac{1}{xy(x+y+z)} = \sum\frac{z}{x+y+z}=1, \]concluding the proof.

parmenides51 wrote: Let $p,q,r \in \mathbb{R}^+$ and for every $n \in \mathbb{N}$ where $pqr=1$ , denote $$ \frac{1}{p^n+q^n+1} + \frac{1}{q^n+r^n+1} + \frac{1}{r^n+p^n+ 1} \leq 1$$ (Art_Ninja) It is an old classic from Baltic Way 2004 and has been posted many times. See e.g. here.