6. Let $a$, $b$, $c$ and $m_{a}$, $m_{b}$, $m_{c}$ represent the lengths of the three sides and the lengths of the medians to these sides of $\triangle ABC$, respectively. Given the following conclusions:
(1) $m_{a}+m_{b}+m_{c} \leq \frac{3}{4}(a+b+c)$;
(4) $m_{a}^{2}+m_{b}^{2}+m_{c}^{2}=\frac{3}{4}\left(a^{2}+b^{2}+c^{2}\right)$.
Among these, the correct ones are ( ).
(A)(1)
(B) (1)(3)
(C)(3)(4)
(D)(2)
6.C.

In Figure 6, let the three medians of $\triangle A B C$ be $A D=m_{a}$,
$$
B E=m_{b}, C F=m_{c},
$$

and they intersect at the centroid $G$.
Then
$$
\begin{array}{l}
G A+G B>A B, \\
G B+G C>B C, \\
G C+G A>C A .
\end{array}
$$

Adding the three inequalities, we get
$$
\begin{array}{l}
2(G A+G B+G C) \\
>A B+B C+C A=a+b+c .
\end{array}
$$

It is easy to see that $G A=\frac{2}{3} m_{a}, G B=\frac{2}{3} m_{b}, G C=\frac{2}{3} m_{c}$.
Therefore, $m_{a}+m_{b}+m_{c}>\frac{3}{4}(a+b+c)$.
Hence (3) is true.
Noting that $m_{a}^{2}=\frac{1}{2}\left(c^{2}+b^{2}\right)-\frac{a^{2}}{4}$,
$$
\begin{array}{l}
m_{b}^{2}=\frac{1}{2}\left(c^{2}+a^{2}\right)-\frac{b^{2}}{4}, \\
m_{c}^{2}=\frac{1}{2}\left(a^{2}+b^{2}\right)-\frac{c^{2}}{4},
\end{array}
$$

Therefore, $m_{a}^{2}+m_{b}^{2}+m_{c}^{2}=\frac{3}{4}\left(a^{2}+b^{2}+c^{2}\right)$. Hence (4) is true.