6. Given real numbers $x, y$ satisfy the condition $3 x^{2}+4 y^{2}=48$, then the maximum value of $\sqrt{x^{2}+y^{2}-4 x+4}+\sqrt{x^{2}+y^{2}-2 x+4 y+5}$ is
6 $8+\sqrt{13}$ Hint:
$$
\begin{aligned}
& \sqrt{x^{2}+y^{2}-4 x+4}+\sqrt{x^{2}+y^{2}-2 x+4 y+5} \\
= & \sqrt{(x-2)^{2}+y^{2}}+\sqrt{(x-1)^{2}+(y+2)^{2}} .
\end{aligned}
$$

Let $P(x, y), A(2,0), B(1,-2)$, then
$$
\sqrt{x^{2}+y^{2}-4 x+4}+\sqrt{x^{2}+y^{2}-2 x+4 y+5}=|P A|+|P B| \text {. }
$$

From the real numbers $x, y$ satisfying the condition $3 x^{2}+4 y^{2}=48$, we know that point $P(x, y)$ lies on the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$, and point $A(2,0)$ is the right focus of the ellipse, while point $B(1,-2)$ is inside the ellipse.
Let the left focus of the ellipse be $F_{1}(-2,0)$, then
$$
|P A|+|P B|=2 a+\left(|P B|-\left|P F_{1}\right|\right) \leqslant 2 a+\left|F_{1} B\right|,
$$

where the equality holds if and only if point $P$ is the intersection point of the ray $B F_{1}$ and the ellipse (where $2 a$ is the length of the major axis of the ellipse).

Therefore, the maximum value of $\sqrt{x^{2}+y^{2}-4 x+4}+\sqrt{x^{2}+y^{2}-2 x+4 y+5}$ is $8+\sqrt{13}$.