USA IMO

First Solution: For any positive number $x$, the quantities $x^2 - 1$ and $x^3 - 1$ have the same sign. Thus, we have $0 \le (x^3 - 1)(x^2 - 1) = x^5 - x^3 - x^2 + 1$, or
$$x^5 - x^2 + 3 \geq x^3 + 2.$$
It follows that
$$(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \geq (a^3 + 2)(b^3 + 2)(c^3 + 2).$$
It suffices to show that
$$(a^3 + 2)(b^3 + 2)(c^3 + 2) \geq (a + b + c)^3. \quad (*)$$
We finish with three approaches.

First approach Expanding both sides of inequality () and cancelling like terms gives
$$\begin{aligned} & a^3b^3c^3 + 3(a^3 + b^3 + c^3) + 2(a^3b^3 + b^3c^3 + c^3a^3) + 8 \\ & \geq 3(a^2b + b^2a + b^2c + c^2b + c^2a + ac^2) + 6abc. \end{aligned}$$
By the AM-GM Inequality, we have $a^3 + a^3 b^3 + 1 \ge 3a^2b$. Combining similar results, the desired inequality reduces to
$$a^3 b^3 c^3 + a^3 + b^3 + c^3 + 1 + 1 \geq 6abc,$$
which is evident by the AM-GM Inequality.

Second Solution: By the AM-GM Inequality,
$$\frac{a^5 + a^5 + 1 + 1 + 1}{5} \geq a^2$$
with equality when $a = 1$. Hence the left hand side of the desired inequality is bounded by
$$\frac{(3a^5 + 12)(3b^5 + 12)(3c^5 + 12)}{125},$$
so it suffices to prove that
$$(3a^5 + 12)(3b^5 + 12)(3c^5 + 12) \geq 125(a + b + c)^3.$$
After expanding the expressions by brute force, we find that this is equivalent to the following, which we will refer to as (†):
$$\begin{aligned} & 27a^5b^5c^5 + 108 \sum_{\text{cyc}} a^5b^5 + 432 \sum_{\text{cyc}} a^5 + 12^3 \\ & \geq 125 \left( \sum_{\text{cyc}} a^3 + 3 \sum_{\text{cyc}} a^2b + 3 \sum_{\text{cyc}} ab^2 + 6abc \right). \end{aligned}$$
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Next, we apply AM-GM seven times to derive the following family of inequalities, referred as (‡),
(1) $a^5 + a^5 + a^5 + 1 + 1 \ge 5a^3$, and so
$$25 \left[ 6 + \sum_{\text{cyc}} 3a^5 \ge 5 \sum_{\text{cyc}} a^3 \right]. \qquad (\ddagger)$$
(2) $a^5 b^5 c^5 + 1 + 1 + 1 + 1 \ge 5abc$, and so
$$27 [a^5 b^5 c^5 + 4 \ge 5abc]. \qquad (\ddagger)$$
(3) $a^5 + a^5 b^5 + 1 + 1 + 1 \ge 5a^2b$, and so
$$54 \left[ \sum_{\text{cyc}} a^5 + \sum_{\text{cyc}} a^5 b^5 + 9 \ge \sum_{\text{cyc}} a^2 b \right]. \qquad (\ddagger)$$
Likewise, we have
$$54 \left[ \sum_{\text{cyc}} b^5 + \sum_{\text{cyc}} a^5 b^5 + 9 \ge \sum_{\text{cyc}} ab^2 \right]. \qquad (\ddagger)$$
(4) $a^5 + b^5 + c^5 + 1 + 1 \ge 5abc$, and so
$$123 [a^5 + b^5 + c^5 + 2 \ge 5abc]. \qquad (\ddagger)$$
(5) $a^5 + a^5 + b^5 + 1 + 1 \ge 5a^2b$, and so
$$21 \left[ \sum_{\text{cyc}} 2a^5 + \sum_{\text{cyc}} b^5 + 6 \ge \sum_{\text{cyc}} a^2 b \right]. \qquad (\ddagger)$$
Likewise, we have
$$21 \left[ \sum_{\text{cyc}} 2b^5 + \sum_{\text{cyc}} a^5 + 6 \ge \sum_{\text{cyc}} ab^2 \right]. \qquad (\ddagger)$$
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Miraculously, the sum of the seven inequalities marked (‡) is precisely inequality (†), which was what we wanted. The equality condition in each application of AM-GM is $a = b = c = 1$, so that is the equality condition for (†) as well.