2. Let $x$, $y$, $z$ all be real numbers, satisfying
$$
\begin{array}{l}
a=-x^{2}+y z-\frac{1}{2011}, \\
b=-y^{2}+x z-\frac{1}{2011}, \\
c=-z^{2}+x y-\frac{1}{2011} .
\end{array}
$$

Then among $a$, $b$, $c$, at least one value ( ).
(A) is greater than 0
(B) is equal to 0
(C) is less than 0
(D) is not greater than 0
2. C.

Notice
$$
\begin{array}{l}
-2(a+b+c) \\
=2 x^{2}-2 y z+2 y^{2}-2 x z+2 z^{2}-2 x y+\frac{6}{2011} \\
=(x-y)^{2}+(y-z)^{2}+(z-x)^{2}+\frac{6}{2011} \\
>0 .
\end{array}
$$

Therefore, $a+b+c<0$.
Hence, at least one of $a$, $b$, and $c$ is less than 0.