6. A point moving in the positive direction of the $O x$ axis has a horizontal coordinate of $x(t)=5(t+1)^{2}+\frac{a}{(t+1)^{5}}$, where $a$ is a positive constant. Find the minimum value of $a$ that satisfies $x(t) \geqslant 24$ for all $t \geqslant 0$.
6. Solution Using the AM-GM inequality $x(t)=5(t+1)^{2}+\frac{a}{2(t+1)^{3}}+\frac{a}{2(t+1)^{5}} \geqslant 7 \sqrt[7]{\frac{a^{2}}{4}}$, equality holds when $(t+1)^{2}=\frac{a}{2(t+1)^{3}}$, thus we have $7 \sqrt[7]{\frac{a^{2}}{4}} \geqslant 24 \Rightarrow a \geqslant 2 \sqrt{\left(\frac{24}{7}\right)^{7}}$, on the other hand, when $a=2 \sqrt{\left(\frac{24}{7}\right)^{7}}, t=\sqrt{\frac{24}{7}}-1$, $x(t)$ $=24$, hence the maximum value of $a$ is $2 \sqrt{\left(\frac{24}{7}\right)^{7}}$.