3. Let $x, y, z \geqslant 0, x+y+z=1$. Then
$$
\sqrt{2011 x+1}+\sqrt{2011 y+1}+\sqrt{2011 z+1}
$$

The sum of the maximum and minimum values is $\qquad$ L.
$$
\begin{array}{l}
\text { 3. } \sqrt{6042}+2 \sqrt{503}+2 \text {. } \\
\text { Let } a=2011 x+1, b=2011 y+1 \text {, } \\
c=2011 z+1 \text {. }
\end{array}
$$

Then \( a, b, c \geqslant 1, a+b+c=2014 \).
First, by the Cauchy-Schwarz inequality, we have
$$
\sqrt{a}+\sqrt{b}+\sqrt{c} \leqslant \sqrt{3(a+b+c)}=\sqrt{6042} \text {. }
$$

Equality holds if and only if \( x=y=z=\frac{1}{3} \).

Second, it is easy to see that for any \( u>v>s>t>0 \), and \( u+t=v+s \), we have
$$
\sqrt{u}+\sqrt{t}<\sqrt{v}+\sqrt{s} .
$$

Therefore, \( \sqrt{a}+\sqrt{b}+\sqrt{c} \geqslant \sqrt{2012}+2 \).
Equality holds if and only if \( x=y=0, z=1 \).