Given $(a+2)^{-\frac{1}{3}}>(1-2 a)^{-\frac{1}{3}}$, find the range of real numbers $a$.
Given the function $y=x^{-\frac{1}{3}}$ is monotonically decreasing in the intervals $(-\infty, 0)$ and $(0,+\infty)$, then
$$
\left\{\begin{array} { l } 
{ a + 2 > 0 , } \\
{ 1 - 2 a > 0 , } \\
{ a + 2 > 0, \\
1-2 a > 0, \\
a < \frac{1}{2}.
\end{array}\right.
$$
Solution 1: The original inequality can be transformed into $\left(\frac{1}{a+2}\right)^{\frac{1}{3}}>\left(\frac{1}{1-2 a}\right)^{\frac{1}{3}}$. Since the function $y=x^{\frac{1}{3}}$ is monotonically increasing in the interval $(-\infty,+\infty)$, then
$$
\frac{1}{a+2}>\frac{1}{1-2 a}
$$

Solving this yields $-2 < a < \frac{1}{2}$.
Solution 2: The original inequality can be transformed into $\left(\frac{1}{a+2}\right)^{\frac{1}{3}}>\left(\frac{1}{1-2 a}\right)^{\frac{1}{3}}$. Since the function $y=x^{\frac{1}{3}}$ is monotonically increasing in the interval $(-\infty,+\infty)$, then
$$
\frac{1}{a+2}>\frac{1}{1-2 a}
$$

Solving this yields $-2 < a < \frac{1}{2}$.
Key Points for Solving: The monotonicity of a function is determined by the exponent. Equivalent transformations of the function form can avoid case-by-case discussions.