3. Given that the area of $\triangle A B C$ is $S, \angle C=\gamma, $ the minimum value of side $C$ opposite to $\angle C$ is
3. $2 \sqrt{S \tan \frac{\gamma}{2}}$.
$c^{2}=a^{2}+b^{2}-2 a b \cos \gamma=(a-b)^{2}+2 a b(1-\cos \gamma)=(a-b)^{2}+\frac{4 S}{\sin \gamma}(1-\cos \gamma)$ (since $S=$ $\left.\frac{1}{2} a b \sin \gamma\right)=(a-b)^{2}+\frac{4 S \cdot 2 \sin ^{2} \frac{r}{2}}{2 \sin \frac{\gamma}{2} \cos \frac{\gamma}{2}}=(a-b)^{2}+4 S \tan \frac{\gamma}{2} \geqslant 4 S \tan \frac{\gamma}{2}$, so $c_{\min}=$ $2 \sqrt{S \tan \frac{\gamma}{2}}$ (when $a=b$, the minimum value is taken).