4.3.12 ** Positive real numbers $a, b, c$ and non-negative real numbers $x, y$ satisfy the condition
$$
a^{6}+b^{6}+c^{6}=3 .(x+1)^{2}+y^{2} \leqslant 2 .
$$

Find the minimum value of $I=\frac{1}{2 a^{3} x+b^{3} y^{2}}+\frac{1}{2 b^{3} x+c^{3} y^{2}}+\frac{1}{2 c^{3} x+a^{3} y^{2}}$.
By the Nesbitt's inequality, we have
$$
\begin{aligned}
I & =\frac{1}{2 a^{3} x+b^{3} y^{2}}+\frac{1}{2 b^{3} x+c^{3} y^{2}}+\frac{1}{2 c^{3} x+a^{3} y^{2}} \\
& \geqslant \frac{9}{2 a^{3} x+b^{3} y^{2}+2 b^{3} x+c^{3} y^{2}+2 c^{3} x+a^{3} y^{2}} \\
& =\frac{9}{2 x\left(a^{3}+b^{3}+c^{3}\right)+y^{2}\left(a^{3}+b^{3}+c^{3}\right)} \quad\left(\left(a^{3}+b^{3}+c^{3}\right)^{2} \leqslant 3\left(a^{6}+b^{6}+c^{6}\right)=9\right) \\
& \geqslant \frac{3}{2 x+y^{2}} \geqslant \frac{3}{2 x+2-(x+1)^{2}} \geqslant \frac{3}{1-x^{2}} \geqslant 3,
\end{aligned}
$$

The equality holds if and only if \(a-b-c=1, x=0, y=1\). Therefore, the minimum value of \(I\) is 3.