, 2010

If $m = 0$, then the inequality is $1 + y^n \ge x^n$. This holds for all positive real numbers $x$ and $y$ iff $n = 0$. Hence $(0, 0)$ is a solution.

Let now both $m$ and $n$ be different from zero. If the pair $(m, n)$ satisfies the condition, then substituting $x$ and $y$ by $\frac{1}{x}$ and $\frac{1}{y}$ we see that the pair $(-m, -n)$ also satisfies the condition. Hence we can assume without loss of generality that $m \ge n$ and $m \ge 0$.

If $m > n$, then by taking $x = 1$ we get $1 + y^n \ge y^m$, which does not hold for $y$ large enough. Hence $m = n$.

By taking $x = y = 4$ we get $2 \cdot 4^m \ge 4^{2m}$ which does not hold for any positive integer $m$. Therefore there are no more suitable pairs.