USA IMO

It suffices to prove that $a_{n+5} \ge a_n^2$ for $n \ge 0$, for then we would have $\sqrt{a_{n+5}} \ge a_n$ and $a_n \ge a_{n-5}^2$ for all $n \ge 5$, which implies the desired result. Adding the inequalities
$$\begin{align*} a_{n+5} &\ge a_{n+4}^2 + \frac{1}{5}, \\ a_{n+4} &\ge a_{n+3}^2 + \frac{1}{5}, \\ a_{n+3} &\ge a_{n+2}^2 + \frac{1}{5}, \\ a_{n+2} &\ge a_{n+1}^2 + \frac{1}{5}, \\ a_{n+1} &\ge a_n^2 + \frac{1}{5} \end{align*}$$
yields
$$\begin{align*} a_{n+5} &\ge \sum_{k=1}^{4} (a_{n+k}^2 - a_{n+k}) + 1 + a_n^2 \\ &= \sum_{k=1}^{4} \left( a_{n+k}^2 - a_{n+k} + \frac{1}{4} \right) + a_n^2 \\ &= \sum_{k=1}^{4} \left( a_{n+k} - \frac{1}{2} \right)^2 + a_n^2 \ge a_n^2, \end{align*}$$
as desired.

(by Alison Miller) For any $k$, we have the following inequality:
$$a_{k+1} + \frac{1}{20} \ge a_k^2 + \frac{1}{5} + \frac{1}{20} = a_k^2 + \frac{1}{4} \ge a_k,$$
because $(a_k - \frac{1}{2})^2 \ge 0$. Summing up this inequalities over $k = n + 1, \dots, n + 4$, we have
$$a_{n+5} + \frac{1}{5} \ge a_{n+1} \ge a_n^2 + \frac{1}{5},$$
and similarly $a_n \ge a_{n-5}^2$, so
$$a_{n+5} \ge a_n^2 \ge a_{n-5}^4,$$
implying the desired result.

Define the function $f(x) = x^2 + \frac{1}{5}$, which is increasing for $x \ge 0$. Also define the function $g(x) = f(x) - x = (x - r_1)(x - r_2)$, where
$$r_1 = \frac{5 - \sqrt{5}}{10} \quad \text{and} \quad r_2 = \frac{5 + \sqrt{5}}{10}.$$
Observe that $0 < r_1 < r_2 < 1$. So, for nonnegative $x$, $g(x)$ is zero for $x = r_1, r_2$, negative for $x \in (r_1, r_2)$, and positive otherwise. Therefore, for nonnegative $x$, $f(x) = x$ for $x = r_1, r_2$, $f(x) < x$ for $x \in (r_1, r_2)$, and $f(x) > x$ otherwise.
Fix $n \ge 5$, and write $a = |a_{n-5}|$. It suffices to prove that $a_{n+5} \ge a_{n-5}^4$. We consider the following cases.
(i) $a \le r_1$. For $x \in [0, r_1]$, observe that $x \le f(x) \le f(r_1) = r_1$. An easy proof by induction on $k = 0, 1, \dots, 9$ then shows that $f^{(k)}(x) \le f^{(k+1)}(x) \le r_1$. Therefore, $a \le f(a) \le f^{(2)}(a) \le \dots \le f^{(10)}(a) = a_{n+5}$. Because $a < 1$, we have $a_{n+5} \ge a > a^4 = a_{n-5}^4$, as desired.
(ii) $a \in (r_2, 1]$. For $x > r_2$, observe that $f(x) > x$ and hence $f(x) > x > r_2$. We can repeat this argument to show that $f^{(k+1)}(x) > f^{(k)}(x) > r_2$ for $k = 0, 1, \dots, 9$. Therefore, $a_{n+5} = f^{(10)}(a) > f^{(9)}(a) > \dots > a \ge a^4 = a_{n-5}^4$, as desired.
(iii) $a > 1$. Notice that for all $k$, $a_{k+1} > a_k^2$. Therefore, $a_{n+5} > a_{n-5}^{10} = a^6 \cdot a_{n-5}^4 > a_{n-5}^4$, as desired.
(iv) $a \in (r_1, r_2)$. For $x \in (r_1, r_2)$, observe that $r_1 = f(r_1) < f(x) < f(r_2) = r_2$. Therefore, $a_{n+5} = f^{(10)}(a) > r_1$. Hence, if we can verify that $r_1 > r_2^4$, then we would have
$$a_{n+5} > r_1 > r_2^4 > a^4 = a_{n-5}^4,$$
as desired. Note that
$$r_2^4 = \frac{\sum_{k=0}^{4} \binom{4}{k} (\sqrt{5})^k 5^{4-k}}{10000},$$
which expands to
$$\frac{5^4 + 4 \cdot 5^3 \cdot \sqrt{5} + 6 \cdot 5^2 \cdot (\sqrt{5})^2 + 4 \cdot 5 \cdot (\sqrt{5})^3 + (\sqrt{5})^4}{10000}.$$
Hence $r_2^4 = (7 + 3\sqrt{5})/50$. Then
$$\begin{aligned} 320 < 324 &\implies 8\sqrt{5} < 18 \\ &\implies 7 + 3\sqrt{5} < 5(5 - \sqrt{5}) \\ &\implies \frac{7 + 3\sqrt{5}}{50} < \frac{5 - \sqrt{5}}{10} \\ &\implies r_2^4 < r_1, \end{aligned}$$
as needed. This completes the proof.