Problem 1420

Let
$$f(b)=b^{3}+b(4+9 a c-8 a-8 c)+a^{3}+c^{3}+4(a+c)-8 a c$$

By the AM-GM inequality, $\left(a^{3}+4 a\right)+\left(c^{3}+4 c\right) \geq 4 a^{2}+4 c^{2} \geq 8 a c$, so the problem is reduced to proving $4+9 a c \geq 8(a+c)$, with equality holding if $a=c=2, b=0$ or $a=b=c=0$. Otherwise, let $x=a+c, y=a c$, then $8 x \geq 9 y+4$, and note that
$$f^{\prime}(b)=3 b^{2}-(8 x-9 y-4) \text {, so } f^{\prime}(b)=0 \Leftrightarrow b=\sqrt{\frac{8 x-9 y-4}{3}}$$

Therefore,
$$f(b) \geq f\left(\sqrt{\frac{8 x-9 y-4}{3}}\right)=\frac{-2}{3 \sqrt{3}}(8 x-9 y-4)^{3 / 2}+x^{3}+4 x-3 x y-8 y=g(y)$$

Since $y \leq \frac{x^{2}}{4}$ and $y \leq \frac{8 x-4}{9}$ (and $x \geq \frac{1}{2}$), we have $y \leq \min \left(\frac{x^{2}}{4}, \frac{8 x-4}{9}\right)=t$
$$g^{\prime}(y)=3 \sqrt{3(8 x-9 y-4)}-(3 x+8) \leq 3 \sqrt{3(8 x-4)}-(3 x+8)<0$$

Thus, $g(y)$ is strictly decreasing. Therefore, $g(y) \geq g(t)$. If $t=\frac{8 x-4}{9}$, then $g(t)=x^{3}+4 x-3 x t-8 t \geq 0$ (or equivalently, $\left.a^{3}+c^{3}+4(a+c)-8 a c \geq 0\right)$ . This only requires considering $t=\frac{x^{2}}{4}$ and $g(t) \geq 0$. Let $s=\frac{x}{2}$, then the inequality $g\left(\frac{x^{2}}{4}\right)=g\left(s^{2}\right) \geq 0$ is equivalent to
$$h(s)=2 s^{3}-8 s^{2}+8 s-\frac{2}{3 \sqrt{3}}\left(16 s-9 s^{2}-4\right)^{3 / 2} \geq 0$$

Since $h^{\prime}(s)=6 s^{2}-16 s+8-(16-18 s) \sqrt{\frac{16 s-9 s^{2}-4}{3}}$, if $h^{\prime}(s)=0$, then we must have $\frac{8}{9} \leq s \leq \frac{8+\sqrt{28}}{9}$ and
$$3\left(2 s^{2}-8 s+4\right)^{2}=(9-8 s)^{2}\left(16 s-9 s^{2}-4\right) \Leftrightarrow(s-1)\left(189 s^{3}-485 s^{2}+372 s-76\right)=0$$

Notice that the equation $189 s^{3}-485 s^{2}+372 s-76=0$ has one real root in the interval $\left[\frac{8}{9}, \frac{8+\sqrt{28}}{9}\right]$, so it is easy to see that $h(s) \geq h(1)=0$. Equality holds if $a=b=c=1$ or $a=b=c=0$ or $a=b=2, c=0$ or any permutation thereof.