$$\begin{array}{l} 12. \text{ Let } f(x)=\frac{1}{\sqrt{1+\mathrm{e}^{-2 x}}}, x \geqslant 0, \text{ then } \mathrm{e}^{-2 x} \leqslant 1, f^{\prime}(x)=\frac{\mathrm{e}^{-2 x}}{\sqrt{\left(1+\mathrm{e}^{-2 x}\right)^{3}}}, f^{\prime \prime}(x)= \\ \frac{\mathrm{e}^{-2 x}}{\sqrt{\left(1+\mathrm{e}^{-2 x}\right)^{3}}}\left(-2+\frac{3 \mathrm{e}^{-2 x}}{1+\mathrm{e}^{-2 x}}\right)=\frac{\mathrm{e}^{-2 x}}{\sqrt{\left(1+\mathrm{e}^{-2 x}\right)^{3}}} \cdot \frac{\mathrm{e}^{-2 x}-2}{1+\mathrm{e}^{-2 x}}0 y=\sec x \text{ has the second derivative } y^{\prime \prime} \\ \frac{1+\sin ^{2} x}{\cos ^{3} x}>0, \text{ so } \\ (\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^{2} \geqslant 6\left(\tan \frac{A+B+C}{}+\sec \frac{A+B+C}{3}\right)=6 \sqrt{3} \end{array}$$

Furthermore
$$\begin{array}{l} -\quad \frac{27}{4}(x+y)(y+z)(z+x) \geqslant(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^{2} \Leftrightarrow \\ \left.-\frac{27}{4} \right| \sec A \geqslant 2\left(\sum \tan A+\sum \sec A\right) \Leftrightarrow \\ \frac{27}{8} \geqslant \sum \cos A \cos B \sin C+\sum \cos A \cos B= \\ \frac{1}{2} \sum \cos A \sin (B+C)+\sum \cos A \cos B= \\ \frac{1}{2} \sum \cos ^{2} A+\sum \cos A \cos B= \\ \frac{1}{2}\left(\sum \cos A\right)^{2} \end{array}$$

Also, since $y=\cos x$ is a concave function on $\left[0, \frac{\pi}{2}\right]$, we have $\cos A+\cos B+\cos C \leqslant$ $3 \cos \frac{A+B+C}{3}=\frac{3 \sqrt{3}}{2}$, thus. $(\cos A+\cos B+\cos C)^{2} \leqslant \frac{27}{4}$. Therefore, the original inequality holds:
$$$$