Romanian Mathematical Olympiad

If $r=0$, the relation is obvious. Otherwise, dividing by $r^{2n}$, one gets the same inequality with $r$ replaced by $\frac{1}{r}$, so one can assume $r\in (0,1]$. Since $r^{2n}+a{r}^n+1\ge r^{2n}-2r^n+1=(1-r^n{)}^2$ and $1-r^n\ge 0$, it is enough to prove that $1-r^n\ge (1-r{)}^n$. This follows from $r^n+(1-r{)}^n\le r+(1-r)=1$.