= Solution
Use the same normalized <spherical cap> area $q$, constant $c_n$ and function $F=\sin^n\rho-q$ as above. Put $\rho_0=\arcsin(1/\sqrt n)$. The supplied ratio estimate gives
$$
c_n<\frac1n\sqrt{\frac{n+1}{2\pi}}\leq\frac{\sqrt{n-1}}n=\sin\rho_0\cos\rho_0.
$$
The middle inequality holds for every $n\geq2$, since $n+1\leq2\pi(n-1)$. Applying the upper projection bound for the <spherical cap> at $\rho_0$ yields
$$
q(\rho_0)\leq c_n\frac{\sin^{n-1}\rho_0}{\cos\rho_0}\leq\sin^n\rho_0,
$$
so $F(\rho_0)\geq0$.
Recall
$$
F'(\rho)=\sin^{n-2}\rho\bigl(n\sin\rho\cos\rho-(n-1)c_n\bigr).
$$
At $\rho_0$ the expression in parentheses is positive, because $n\sin\rho_0\cos\rho_0=\sqrt{n-1}$ and $(n-1)c_n<\sqrt{n-1}$. It increases up to $\pi/4$ and then decreases strictly to a negative value at $\pi/2$. Hence it crosses zero exactly once on $[\rho_0,\pi/2]$, from positive to negative. The minimum of $F$ is again an endpoint value; $F(\rho_0)\geq0$ and $F(\pi/2)=1/2$. \b[This proves the stronger range of the <spherical cap area upper bound>:]
$$
\boxed{\operatorname{Vol}_{n-1}(D_\rho)\leq\sin^n\rho\,s_{n-1}\qquad(\sin\rho\geq n^{-1/2}).}
$$
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