APMO

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

We prove the stronger inequality $$\begin{array}{c}\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)\ge 3(a+b+c{)}^2,\end{array}\text{\hspace{1em}(*)}$$ which implies the proposed inequality because $(a+b+c{)}^2\ge 3(ab+bc+ca)$ is equivalent to $(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\ge 0$, which is immediate. The inequality $(\ast )$ is equivalent to $$\left((b^2+2)(c^2+2)-3\right)a^2-6(b+c)a+2(b^2+2)(c^2+2)-3(b+c{)}^2\ge 0.$$ 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 $$(3(b+c){)}^2-\left((b^2+2)(c^2+2)-3\right)\left(2(b^2+2)(c^2+2)-3(b+c{)}^2\right)\le 0.$$ This simplifies to $$\begin{array}{c}-2(b^2+2)(c^2+2)+3(b+c{)}^2+6\le 0.\end{array}\text{\hspace{1em}(**)}$$ Now we look at $(\ast \ast )$ as a quadratic inequality in $b$ with negative leading coefficient $-2c^2-1$: $$(-2c^2-1)b^2+6cb-c^2-2\le 0.$$ It suffices to show that the discriminant of $(\ast \ast )$ is non-positive, which is equivalent to $$9c^2-(2c^2+1)(c^2+2)\le 0$$ It simplifies to $-2(c^2-1{)}^2\le 0$, 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 $$4{\sec }^{2}A{\sec }^{2}B{\sec }^{2}C\ge 9(\tan A\tan B+\tan B\tan C+\tan C\tan A).$$ Substituting $\sec x=\frac{1}{\cos x}$ for $x\in \{A,B,C\}$ and clearing denominators, the inequality is equivalent to $$\cos A\cos B\cos C(\sin A\sin B\cos C+\cos A\sin B\sin C+\sin A\cos B\sin C)\le \frac{4}{9}$$ Since $$\begin{array}{rl}& \cos (A+B+C)=\cos A\cos (B+C)-\sin A\sin (B+C)\\ =& \cos A\cos B\cos C-\cos A\sin B\sin C-\sin A\cos B\sin C-\sin A\sin B\cos C,\end{array}$$ we rewrite our inequality as $$\cos A\cos B\cos C(\cos A\cos B\cos C-\cos (A+B+C))\le \frac{4}{9}$$ 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, $$\cos A\cos B\cos C\le {\left(\frac{\cos A+\cos B+\cos C}{3}\right)}^3\le {\cos }^3\frac{A+B+C}{3}={\cos }^3\theta$$ 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 $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$, $\cos A\cos B\cos C(\cos A\cos B\cos C-\cos (A+B+C))\le {\cos }^3\theta ({\cos }^3\theta -\cos 3\theta )=3{\cos }^4\theta (1-{\cos }^2\theta )$. Finally, by AM-GM (notice that $1-{\cos }^2\theta ={\sin }^2\theta >0$), $3{\cos }^4\theta (1-{\cos }^2\theta )=\frac{3}{2}{\cos }^2\theta \cdot {\cos }^2\theta (2-2{\cos }^2\theta )\le \frac{3}{2}{\left(\frac{{\cos }^2\theta +{\cos }^2\theta +(2-2{\cos }^2\theta )}{3}\right)}^3=\frac{4}{9}$,

and the result follows.