First we prove that
$$\frac{a^2}{\sqrt{a^4 + 4b + 4c}} + \frac{b^2}{\sqrt{b^4 + 4c + 4a}} + \frac{c^2}{\sqrt{c^4 + 4a + 4b}} \ge 1$$
Applying AM-GM inequality, we conclude that
$$\begin{aligned} \frac{a^2}{\sqrt{a^4 + 4b + 4c}} &= \frac{a^2}{\sqrt{a^4 + 4ab^2c + 4abc^2}} \\ &\ge \frac{a^2}{\sqrt{a^4 + (a^2b^2 + a^2c^2 + b^2c^2 + b^4) + (a^2b^2 + a^2c^2 + b^2c^2 + c^4)}} \\ &= \frac{a^2}{a^2 + b^2 + c^2} \end{aligned}$$
Similarly,
$$\frac{b^2}{\sqrt{b^4 + 4c + 4a}} \ge \frac{b^2}{b^2 + c^2 + a^2}; \quad \frac{c^2}{\sqrt{c^4 + 4a + 4b}} \ge \frac{c^2}{c^2 + a^2 + b^2}$$
Therefore
$$\frac{a^2}{\sqrt{a^4 + 4b + 4c}} + \frac{b^2}{\sqrt{b^4 + 4c + 4a}} + \frac{c^2}{\sqrt{c^4 + 4a + 4b}} \ge 1$$

Now it remains to prove that
$$\frac{1}{\sqrt{a^3 + 2b^3 + 6}} + \frac{1}{\sqrt{b^3 + 2c^3 + 6}} + \frac{1}{\sqrt{c^3 + 2a^3 + 6}} \le 1$$
Applying the AM-GM inequality we conclude that
$$\begin{aligned} \frac{1}{\sqrt{a^3 + 2b^3 + 6}} &\le \frac{1}{\sqrt{3ab^2 + 6}} \\ &= \sqrt{\frac{abc}{3ab^2 + 6abc}} = \sqrt{\frac{1}{3}\left(\frac{c}{b+2c}\right)} \\ &\le \frac{1}{2}\left(\frac{1}{3} + \frac{c}{b+2c}\right) \\ &= \frac{1}{4}\left(\frac{5}{3} - \frac{a}{a+2b}\right). \end{aligned}$$

Therefore using the Cauchy-Schwartz inequality we have
$$\begin{aligned} & \frac{1}{\sqrt{a^3 + 2b^3 + 6}} + \frac{1}{\sqrt{b^3 + 2c^3 + 6}} + \frac{1}{\sqrt{c^3 + 2a^3 + 6}} \\ \le & \frac{5}{4} - \frac{1}{4} \left( \frac{b}{b+2c} + \frac{c}{c+2a} + \frac{a}{a+2b} \right) \\ \le & \frac{5}{4} - \frac{(a+b+c)^2}{4(b(b+2c) + c(c+2a) + a(a+2b))} = 1. \end{aligned}$$
Hence the proof of this problem follows. □