Example 2 The solution to the functional inequality $2 f(x)+x^{2} f\left(\frac{1}{x}\right) \geqslant \frac{3 x^{3}-x^{2}+4 x+3}{x+1}$ is
The given functional inequality imposes no constraints on $x=0,-1$, so $f(x)$ can take any value at these points. The complete solution is as follows: when $x \neq 0,-1$, let
$$
2 f(x)+x^{2} f\left(\frac{1}{x}\right)=\frac{3 x^{3}-x^{2}+4 x+3}{x+1}+g(x), \forall g(x) \geqslant 0.
$$

Let $x=\frac{1}{y}$, then we get $f(y)+2 y^{2} f\left(\frac{1}{y}\right)=\frac{3 y^{3}+4 y^{2}-y+3}{y+1}+g\left(\frac{1}{y}\right)$. Changing $y$ to $x$ gives
$$
f(x)+2 x^{2} f\left(\frac{1}{x}\right)=\frac{3 x^{3}+4 x^{2}-x+3}{x+1}+x^{2} g\left(\frac{1}{x}\right).
$$
$2 \times$ (1) - (2) eliminates $f\left(\frac{1}{x}\right)$, yielding
$$
f(x)=x^{2}-3 x+6-\frac{5}{x+1}+\frac{2}{3} g(x)-\frac{x^{2}}{3} g\left(\frac{1}{x}\right), \forall g(x) \geqslant 0, x \neq 0,1.
$$

When $x=0,-1$, the given functional inequality imposes no constraints on $f(x)$, so $f(x)$ can take any value or be undefined at these points.

Verification Substituting (3) into $2 f(x)+x^{2} f\left(\frac{1}{x}\right)-\frac{3 x^{3}-x^{2}+4 x+3}{x+1}=g(x) \geqslant 0$, we see that the function $f(x)$ given by (3) is a solution to the given functional inequality.

Explanation Taking $g(x)=0$, we obtain the solution to the corresponding functional equation $2 f(x)+x^{2} f\left(\frac{1}{x}\right)=\frac{3 x^{3}-x^{2}+4 x+3}{x+1}$ of the given functional inequality, which is (where $b, c$ are arbitrary constants):
(1) $f(x)=x^{2}-3 x+6-\frac{5}{x+1}(x \neq 0,-1)$;
(2) $f(x)=\left\{\begin{array}{l}x^{2}-3 x+6-\frac{5}{x+1}, x \neq 0,-1 \\ c, x=-1\end{array}\right.$;
(3) $f(x)=\left\{\begin{array}{l}x^{2}-3 x+6-\frac{5}{x+1}, x \neq 0,-1 \\ b, x=0\end{array}\right.$;
(4) $f(x)=\left\{\begin{array}{l}b, x=0 \\ x^{2}-3 x+6-\frac{5}{x+1}, x \neq 0,-1 . \\ c, x=-1\end{array}\right.$

Clearly, they are also particular solutions to the given functional inequality.