5. In $\mathrm{V} A B C$, $A, B, C$ represent the three interior angles, $\frac{1}{\sin ^{2} A}+\frac{1}{\sin ^{2} B}+\frac{4}{1+\sin C}$ The minimum value is $\qquad$
5. $16-8 \sqrt{2}$

Analysis: $\frac{1}{\sin ^{2} A}+\frac{1}{\sin ^{2} B}+\frac{4}{1+\sin C} \geq \frac{2}{\sin A \sin B}+\frac{4}{1+\sin C}=\frac{4}{\cos (A-B)-\cos (A+B)}+\frac{4}{1+\sin C}$ $\geq \frac{4}{1+\cos C}+\frac{4}{1+\sin C}$, by Cauchy-Schwarz inequality $\frac{4}{1+\cos C}+\frac{4}{1+\sin C} \geq \frac{16}{2+\cos C+\sin C}$ $\geq \frac{16}{2+\sqrt{2} \sin (C+\pi / 4)} \geq \frac{16}{2+\sqrt{2}}=16-8 \sqrt{2}$

Equality holds if and only if $A=B=\frac{3 \pi}{8}, C=\frac{\pi}{4}$, hence the minimum value of $f$ is $16-8 \sqrt{2}$.