Example 2 The abscissa of a point moving in the positive direction of the $O x$ axis is $x(t)=5(t+1)^{2}+\frac{a}{(t+1)^{5}}$, where $a$ is a positive constant. Find the minimum value of $a$ that satisfies $x(t) \geqslant 24$ for all $t \geqslant 0$.
Solution: Using the arithmetic-geometric mean inequality, we get
$$
\begin{aligned}
x(t)= & (\underbrace{t+1)^{2}+\cdots+(t+1)^{2}}_{5 \uparrow} \\
& +\frac{a}{2(t+1)^{5}}+\frac{a}{2(t+1)^{5}} \\
\geqslant & 7 \sqrt[7]{\frac{a^{2}}{4}} .
\end{aligned}
$$

Equality holds if and only if $(t+1)^{2}=\frac{a}{2(t+1)^{5}}$.
Thus, we have $7 \sqrt[7]{\frac{a^{2}}{4}} \geqslant 24$, which implies $a \geqslant 2 \sqrt{\left(\frac{24}{7}\right)^{7}}$.
On the other hand, when $a=2 \sqrt{\left(\frac{24}{7}\right)^{7}}, t=\sqrt{\frac{24}{7}}-1$, we have $x(t)=24$, so the minimum value of $a$ is $2 \sqrt{\left(\frac{24}{7}\right)^{7}}$.