Notice that $a^3 + b^3 + 1 \ge 3ab$ and $b^3 + 1 + 1 \ge 3b$ to obtain that $\frac{1}{a^3+2b^3+6} \le \frac{1}{3ab+3b+3}$, hence it is sufficient to show that $\frac{1}{ab+b+1} + \frac{1}{bc+c+1} + \frac{1}{ca+a+1} \le 1$.

To this end, observe that $\frac{1}{ab+b+1} + \frac{1}{bc+c+1} + \frac{1}{ca+a+1} = \frac{1}{ab+b+1} + \frac{ab}{ab^2c+abc+ab} + \frac{b}{abc+ab+b} \stackrel{abc \ge 1}{\le} \frac{1}{ab+b+1} + \frac{ab}{b+1+ab} + \frac{b}{1+ab+b} = \frac{1+ab+b}{ab+b+1} = 1$.

Alternative Solution:

Subtract $1/6$ from each of the left hand-side summands and write successively $\sum_{cyc} \left(\frac{1}{a^3+2b^3+6} - \frac{1}{6}\right) \le \frac{1}{3} - \frac{1}{2}$, then $\sum_{cyc} \frac{-a^3-2b^3}{6(a^3+2b^3+6)} \le -\frac{1}{6}$ or further $\sum_{cyc} \frac{a^3+2b^3}{a^3+2b^3+6} \ge 1$.

To this end, notice that
$$\begin{aligned} \sum_{cyc} \frac{a^3}{a^3 + 2b^3 + 6} &= \sum_{cyc} \frac{a^4}{a^4 + 2ab^3 + 6a} \\ &\stackrel{CBS}{\ge} \frac{(a^2 + b^2 + c^2)^2}{a^4 + b^4 + c^4 + 2(ab^3 + bc^3 + ca^3) + 6(a + b + c)} \stackrel{(1)}{\ge} \frac{1}{3}. \end{aligned}$$

The inequality (1) rewrites $3(a^4+b^4+c^4)+6(a^2b^2+b^2c^2+c^2a^2) \ge a^4+b^4+c^4+2(ab^3+bc^3+ca^3)+6(a+b+c)$, and follows from $2(a^4+b^4+c^4) \ge 2(ab^3+bc^3+ca^3)$ and $6(a^2b^2+b^2c^2+c^2a^2) \ge 6(ab \cdot bc+bc \cdot ca+ca \cdot ab) = 6abc(a+b+c) \ge 6(a+b+c)$.

On the other hand,
$$\begin{aligned} \sum_{cyc} \frac{b^3}{a^3 + 2b^3 + 6} &= \sum_{cyc} \frac{b^4}{ba^3 + 2b^4 + 6b} \\ &\stackrel{CBS}{\ge} \frac{(a^2 + b^2 + c^2)^2}{2(a^4 + b^4 + c^4) + (ba^3 + cb^3 + ac^3) + 6(a+b+c)} \\ &\stackrel{(2)}{\ge} \frac{1}{3}. \end{aligned}$$

The inequality (2) rewrites $3(a^4+b^4+c^4)+6(a^2b^2+b^2c^2+c^2a^2) \ge 2(a^4+b^4+c^4)+(ba^3+cb^3+ac^3)+6(a+b+c)$, and follows from $a^4+b^4+c^4 \ge ba^3+cb^3+ac^3$ and $6(a^2b^2+b^2c^2+c^2a^2) \ge 6(ab \cdot bc+bc \cdot ca+ca \cdot ab) = 6abc(a+b+c) \ge 6(a+b+c)$.

Alternative Solution:

As $a^3 + 2b^3 = a^3 + b^3 + b^3 \ge 3ab^2$, it suffices to prove that $\frac{1}{ab^2+2} + \frac{1}{bc^2+2} + \frac{1}{ca^2+2} \le 1$. Rewrite the inequality as $-\frac{ab^2}{2(ab^2+2)} - \frac{bc^2}{2(bc^2+2)} - \frac{ca^2}{2(ca^2+2)} \le 1 - \frac{3}{2}$, or, equivalently,
$$\frac{ab^2}{ab^2+2} + \frac{bc^2}{bc^2+2} + \frac{ca^2}{ca^2+2} \ge 1.$$
Recall that $abc \ge 1$, and write $\frac{ab^2}{ab^2+2abc} + \frac{bc^2}{bc^2+2abc} + \frac{ca^2}{ca^2+2abc} \ge 1$ to observe that it is enough to show that $\frac{b}{b+2c} + \frac{c}{c+2a} + \frac{a}{a+2b} \ge 1$. Indeed, $\frac{b}{b+2c} + \frac{c}{c+2a} + \frac{a}{a+2b} = \frac{b^2}{b^2+2bc} + \frac{c^2}{c^2+2ca} + \frac{a^2}{a^2+2ab} \stackrel{CBS}{\ge} \frac{(a+b+c)^2}{b^2+2bc+c^2+2ca+a^2+2ab} = 1$, which concludes the proof.