* 3. $A, B, C$ are the three interior angles of $\triangle A B C$, and $\operatorname{ctg} \frac{A}{2}$ $+\operatorname{ctg} \frac{B}{2}+\operatorname{ctg} \frac{C}{2}-2(\operatorname{ctg} A+\operatorname{ctg} B+\operatorname{ctg} C) \geqslant T$. Then $T_{\max }=$ $\qquad$
3. $\sqrt{3}$. From $\operatorname{ctg} \frac{A}{2}-2 \operatorname{ctg} A=\operatorname{tg} \frac{A}{2}$,

we have $\operatorname{tg} \frac{A}{2}+\operatorname{tg} \frac{B}{2}+\operatorname{tg} \frac{C}{2} \geqslant T$.
$$
\begin{array}{l}
\text { (tg } \left.\frac{A}{2}+\operatorname{tg} \frac{B}{2}+\operatorname{tg} \frac{C}{2}\right)^{2} \\
=\operatorname{tg}^{2} \frac{A}{2}+\operatorname{tg}^{2} \frac{B}{2}+\operatorname{tg}^{2} \frac{C}{2} \\
\quad+2 \operatorname{tg} \frac{A}{2} \operatorname{tg} \frac{B}{2}+2 \operatorname{tg} \frac{B}{2} \operatorname{tg} \frac{C}{2}+2 \operatorname{tg} \frac{C}{2} \operatorname{tg} \frac{A}{2} \\
\geqslant 3\left(\operatorname{tg} \frac{A}{2} \operatorname{tg} \frac{B}{2}+\operatorname{tg} \frac{B}{2} \operatorname{tg} \frac{C}{2}+\operatorname{tg} \frac{C}{2} \operatorname{tg} \frac{A}{2}\right) \\
=3 .
\end{array}
$$

Thus, $T_{\max }=\sqrt{3}$.