11. Given that for all real numbers $x$, we have
$$
|x+1|+\sqrt{x-1} \geqslant m-|x-2|
$$

always holds. Then the maximum value that $m$ can take is $\qquad$ .
When $-1 \leqslant x \leqslant 2$, the minimum value of $|x+1|+|x-2|$ is 
3. Since $\sqrt{x-1} \geqslant 0$, when $x=1$, 
$$
|x+1|+\sqrt{x-1}+|x-2|
$$

the minimum value is 3, so, $3 \geqslant m$, hence the maximum value of $m$ is 3.