China Hong Kong Mathematical Olympiad

Let $f(a, b, c) = ab^2 + bc^2 + ca^2$. Note that it suffices to show
$$f(a, b, c) - f(a, c, b) \le 0, \qquad (1)$$
$$f(a, b, c) + f(a, c, b) \le \frac{27}{4}, \qquad (2)$$
since adding these yields the result. To prove (1), observe that
$$f(a, b, c) - f(a, c, b) = (a - b)(b - c)(c - a) \le 0.$$
To prove (2), by homogenizing the inequality, we need to prove
$$ab^2 + bc^2 + ca^2 + a^2b + b^2c + c^2a \le \frac{1}{4}(a + b + c)^3.$$
This is equivalent to
$$a^3 + b^3 + c^3 + 6abc \ge ab^2 + bc^2 + ca^2 + a^2b + b^2c + c^2a,$$
which holds by Schur's inequality and $3abc \ge 0$. Equality holds when $c = 0$ and $a = b = \frac{3}{2}$.