9. Let the function $f(x)=5(x+1)^{2}+\frac{a}{(x+1)^{5}}(a>0)$, when $x \geqslant 0, f(x) \geqslant 24$, the minimum value of $a$ is
9. $2 \sqrt{\left(\frac{24}{7}\right)^{7}}$. Since $f(x)=\underbrace{(x+1)^{2}+\cdots+(x+1)^{2}}_{1 \uparrow}+\frac{a}{2(x+1)^{5}}+\frac{a}{2(x+1)^{5}} \geqslant 7 \cdot \sqrt[7]{\frac{a^{2}}{4}}($ when and only when $(x+1)^{2}=\frac{a}{2(x+1)^{3}}$, the equality holds $)$.
Thus, $7 \cdot \sqrt[7]{\frac{a^{2}}{4}} \geqslant 24 \Rightarrow a \geqslant 2 \sqrt{\left(\frac{24}{7}\right)^{7}}$.
On the other hand, when $a=2 \sqrt{\left(\frac{24}{7}\right)^{7}}$, $x=\sqrt{\frac{24}{7}}-1, f(x)=24$.