$$
x(t)=5(t+1)^{2}+\frac{a}{(t+1)^{5}},
$$

where $a$ is a positive constant. Find the minimum value of $a$ that satisfies $x(t) \geqslant 24$ for all $t \geqslant 0$.
By using the AM-GM inequality, we have
$$
x(t)=\underbrace{(t+1)^{2}+\cdots+(t+1)^{2}}_{5 \uparrow}+\frac{a}{2(t+1)^{5}}+\frac{a}{2(t+1)^{5}} \geqslant 7 \sqrt[7]{\frac{a^{2}}{4}},
$$

and the equality holds when $(t+1)^{2}=\frac{a}{2(t+1)^{5}}$. Therefore, 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 $x(t)=24$.
Thus, the minimum value of $a$ is $2 \sqrt{\left(\frac{24}{7}\right)^{7}}$.
Note: If we take special values of $t$, for example, $t=0$ gives $a \geqslant 19$, and $t=1$ gives $a \geqslant 128$, etc.