6. Let $a, b, c$ be the lengths of the three sides of $\triangle ABC$, and $t_{a}, t_{b}, t_{c}$ be the lengths of the corresponding angle bisectors of $\triangle ABC$. Given the following conclusions:
(1) $t_{a} t_{b} t_{c} \geqslant a b c$,
(2) $t_{a} t_{b} t_{c}<\frac{\sqrt{3}}{2} a b c$,
(3) $t_{a}^{2}+t_{b}^{2}+t_{c}^{2} \geqslant a^{2}+b^{2}+c^{2}$,
(4) $t_{a}^{2}+t_{b}^{2}+t_{c}^{2}<a^{2}+b^{2}+c^{2}$,
(5) $t_{a}+t_{b}+t_{c} \geqslant a+b+c$,
(6) $t_{a}+t_{b}+t_{c}<a+b+c$.

Among them, the correct ones are ( ).
(A)(1)(3)(5)
(B)(1)(3)(6)
(C)(2)(4)(5)
(D)(2)(4)(6)
6. D.

Notice,
$$
\begin{aligned}
t_{a} & =\frac{b c \sin A}{(b+c) \sin \frac{A}{2}}=\frac{2 b c \cos \frac{A}{2}}{b+c}, \\
t_{b} & =\frac{2 a c \cos \frac{B}{2}}{a+c}, \\
t_{c} & =\frac{2 a b \cos \frac{C}{2}}{a+b} .
\end{aligned}
$$

Assume without loss of generality that $a \geqslant b \geqslant c$. Then
$$
\angle A \geqslant \frac{\pi}{3} \Rightarrow \cos \frac{A}{2} \leqslant \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2} \text {. }
$$

Also, $\cos \frac{B}{2}<1, \cos \frac{C}{2}<1$, so
$$
\begin{array}{l}
t_{a} t_{b} t_{c}<\frac{\sqrt{3}}{2} \cdot \frac{2 b c}{b+c} \cdot \frac{2 a c}{a+c} \cdot \frac{2 a b}{a+b} \\
\leqslant \frac{\sqrt{3}}{2} \cdot \sqrt{b c} \cdot \sqrt{a c} \cdot \sqrt{a b}=\frac{\sqrt{3}}{2} a b c .
\end{array}
$$

Therefore, conclusion (1) is incorrect, and conclusion (2) is correct.
$=\frac{2 b c \cos \frac{A}{2}}{b+c}<\frac{2 b c}{b+c} \leqslant \sqrt{b c}$, similarly,
$$
\begin{array}{l}
t_{b}<\sqrt{a c}, t_{c}<\sqrt{a b} . \\
\text { Hence } t_{a}^{2}+t_{b}^{2}+t_{c}^{2}<b c+a c+a b \\
\leqslant a^{2}+b^{2}+c^{2} .
\end{array}
$$

Therefore, conclusion (3) is incorrect, and conclusion (4) is correct.
Similarly, $t_{a}+t_{b}+t_{c}<\sqrt{b c}+\sqrt{a c}+\sqrt{a b}$ $\leqslant a+b+c$.
Therefore, conclusion (5) is incorrect, and conclusion (6) is correct.