Irish Mathematical Olympiad

To prove
$$x_{n+1} = \sqrt[n+1]{9^n(n+10)} < \sqrt[n]{9^{n-1}(n+9)} = x_n \quad \text{for all } n \ge 2$$
is equivalent to showing
$$(9^n(n+10))^n < (9^{n-1}(n+9))^{n+1} \quad \text{for } n=2,3,\dots$$
Simplifying this, the claim is that
$$9(n + 10)^n < (n + 9)^{n+1}, \quad n = 2, 3, \dots$$
But, by the AM-GM inequality,
$$\sqrt[n+1]{9(n+10)^n} \le \frac{9+n(n+10)}{n+1} = n+9$$
Hence, $9(n+10)^n \le (n+9)^{n+1}$ for all $n \ge 2$, and so, in the first place, the sequence is decreasing. Moreover, since $9 \ne n+10$ for any $n \ge 0$, the AM-GM inequality is strict. Hence, the sequence is strictly decreasing.

Next,
$$x_n = \sqrt[n]{9^{n-1}(n+9)} > 9, \quad n = 2, 3, \dots$$
so $x_n$ is bounded below by $9$. The result follows.