Asian Pacific Mathematics Olympiad

Let $p = a + b + c$, $q = ab + bc + ca$, and $r = abc$. The inequality simplifies to
$$a^{2}b^{2}c^{2} + 2(a^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2}) + 4(a^{2} + b^{2} + c^{2}) + 8 - 9(ab + bc + ca) \geq 0.$$
Since $a^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2} = q^{2} - 2pr$ and $a^{2} + b^{2} + c^{2} = p^{2} - 2q$,
$$r^{2} + 2q^{2} - 4pr + 4p^{2} - 8q + 8 - 9q \geq 0,$$
which simplifies to
$$\begin{equation*} r^{2} + 2q^{2} + 4p^{2} - 17q - 4pr + 8 \geq 0. \tag{I} \end{equation*}$$
Bearing in mind that equality occurs for $a = b = c = 1$, which means that, for instance, $p = 3r$, one can rewrite (I) as
$$\begin{equation*} \left(r - \frac{p}{3}\right)^{2} - \frac{10}{3}pr + \frac{35}{9}p^{2} + 2q^{2} - 17q + 8 \geq 0. \tag{II} \end{equation*}$$
Since $(ab - bc)^{2} + (bc - ca)^{2} + (ca - ab)^{2} \geq 0$ is equivalent to $q^{2} \geq 3pr$, rewrite (II) as
$$\begin{equation*} \left(r - \frac{p}{3}\right)^{2} + \frac{10}{9}(q^{2} - 3pr) + \frac{35}{9}p^{2} + \frac{8}{9}q^{2} - 17q + 8 \geq 0. \tag{III} \end{equation*}$$
Finally, $a = b = c = 1$ implies $q = 3$; then rewrite (III) as
$$\left(r - \frac{p}{3}\right)^{2} + \frac{10}{9}(q^{2} - 3pr) + \frac{35}{9}(p^{2} - 3q) + \frac{8}{9}(q - 3)^{2} \geq 0.$$
This final inequality is true because $q^{2} \geq 3pr$ and $p^{2} - 3q = \frac{1}{2}[(a - b)^{2} + (b - c)^{2} + (c - a)^{2}] \geq 0$.

We prove the stronger inequality
$$\begin{equation*} \left(a^{2} + 2\right)\left(b^{2} + 2\right)\left(c^{2} + 2\right) \geq 3(a + b + c)^{2}, \tag{*} \end{equation*}$$
which implies the proposed inequality because $(a + b + c)^{2} \geq 3(ab + bc + ca)$ is equivalent to $(a - b)^{2} + (b - c)^{2} + (c - a)^{2} \geq 0$, which is immediate.
The inequality $(*)$ is equivalent to
$$\left((b^{2} + 2)(c^{2} + 2) - 3\right)a^{2} - 6(b + c)a + 2(b^{2} + 2)(c^{2} + 2) - 3(b + c)^{2} \geq 0.$$
Seeing this inequality as a quadratic inequality in $a$ with positive leading coefficient $(b^{2} + 2)(c^{2} + 2) - 3 = b^{2}c^{2} + 2b^{2} + 2c^{2} + 1$, it suffices to prove that its discriminant is non-positive, which is equivalent to
$$(3(b + c))^{2} - \left((b^{2} + 2)(c^{2} + 2) - 3\right)\left(2(b^{2} + 2)(c^{2} + 2) - 3(b + c)^{2}\right) \leq 0.$$
This simplifies to
$$\begin{equation*} -2(b^{2} + 2)(c^{2} + 2) + 3(b + c)^{2} + 6 \leq 0. \tag{**} \end{equation*}$$
Now we look at $(**)$ as a quadratic inequality in $b$ with negative leading coefficient $-2c^{2} - 1$:
$$(-2c^{2} - 1)b^{2} + 6cb - c^{2} - 2 \leq 0.$$
It suffices to show that the discriminant of $(**)$ is non-positive, which is equivalent to
$$9c^{2} - (2c^{2} + 1)(c^{2} + 2) \leq 0$$
It simplifies to $-2(c^{2} - 1)^{2} \leq 0$, which is true. The equality occurs for $c^{2} = 1$, that is, $c = 1$, for which $b = \frac{6c}{2(2c^{2} + 1)} = 1$, and $a = \frac{6(b + c)}{2((b^{2} + 2)(c^{2} + 2) - 3)} = 1$.

Let $A, B, C$ be angles in $(0, \pi/2)$ such that $a = \sqrt{2} \tan A$, $b = \sqrt{2} \tan B$, and $c = \sqrt{2} \tan C$. Then the inequality is equivalent to
$$4 \sec^{2} A \sec^{2} B \sec^{2} C \geq 9(\tan A \tan B + \tan B \tan C + \tan C \tan A).$$
Substituting $\sec x = \frac{1}{\cos x}$ for $x \in \{A, B, C\}$ and clearing denominators, the inequality is equivalent to
$$\cos A \cos B \cos C (\sin A \sin B \cos C + \cos A \sin B \sin C + \sin A \cos B \sin C) \leq \frac{4}{9}$$
Since
$$\begin{aligned} & \cos (A + B + C) = \cos A \cos (B + C) - \sin A \sin (B + C) \\ = & \cos A \cos B \cos C - \cos A \sin B \sin C - \sin A \cos B \sin C - \sin A \sin B \cos C, \end{aligned}$$
we rewrite our inequality as
$$\cos A \cos B \cos C (\cos A \cos B \cos C - \cos (A + B + C)) \leq \frac{4}{9}$$
The cosine function is concave down on $(0, \pi/2)$. Therefore, if $\theta = \frac{A + B + C}{3}$, by the AM-GM inequality and Jensen's inequality,
$$\cos A \cos B \cos C \leq \left(\frac{\cos A + \cos B + \cos C}{3}\right)^{3} \leq \cos^{3} \frac{A + B + C}{3} = \cos^{3} \theta$$
Therefore, since $\cos A \cos B \cos C - \cos (A + B + C) = \sin A \sin B \cos C + \cos A \sin B \sin C + \sin A \cos B \sin C > 0$, and recalling that $\cos 3\theta = 4\cos^{3} \theta - 3\cos \theta$,
$\cos A \cos B \cos C (\cos A \cos B \cos C - \cos (A + B + C)) \leq \cos^{3} \theta (\cos^{3} \theta - \cos 3\theta) = 3\cos^{4} \theta (1 - \cos^{2} \theta)$.
Finally, by AM-GM (notice that $1 - \cos^{2} \theta = \sin^{2} \theta > 0$),
$3\cos^{4} \theta (1 - \cos^{2} \theta) = \frac{3}{2} \cos^{2} \theta \cdot \cos^{2} \theta (2 - 2\cos^{2} \theta) \leq \frac{3}{2} \left(\frac{\cos^{2} \theta + \cos^{2} \theta + (2 - 2\cos^{2} \theta)}{3}\right)^{3} = \frac{4}{9}$,

and the result follows.