Let us make such transformation:
$$1 = \frac{ab+bc+ca}{3} \ge \sqrt[3]{(abc)^2} \Leftrightarrow abc \le a \le \frac{1}{bc},\ b \le \frac{1}{ac},\ c \le \frac{1}{ab},$$
hence $a+b+c \le \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$. Then we use well-known inequality:
$$\frac{a_1^2}{b_1} + \frac{a_2^2}{b_2} + \dots + \frac{a_n^2}{b_n} \ge \frac{(a_1+a_2+\dots+a_n)^2}{b_1+b_2+\dots+b_n}.$$
Then we will make such transformation:
$$\begin{align*} \frac{1}{2a^3+1} + \frac{1}{2b^3+1} + \frac{1}{2c^3+1} &= \frac{\frac{1}{a^2}}{2a+\frac{1}{a^2}} + \frac{\frac{1}{b^2}}{2b+\frac{1}{b^2}} + \frac{\frac{1}{c^2}}{2c+\frac{1}{c^2}} \\ &\ge \frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2}{2a+2b+2c+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}} \\ &\ge \frac{\frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2}{2}}{\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}} \\ &= \frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2}{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2} = 1, \end{align*}$$
That is what we had to prove.