3. If $a, b, c$ are real numbers, $x=a^{2}-2 b+\frac{\pi}{3}, y=$ $b^{2}-2 c+\frac{\pi}{6}, z=c^{2}-2 a+\frac{\pi}{2}$, then among $x, y, z$
(A) at least one is greater than 0
(B) at least one is not less than 0
(C) at least one is not greater than 0
(D) at least one is less than 0
3. (A).
$$
\begin{array}{l} 
x+y+z \\
=\left(a^{2}-2 b+\frac{\pi}{2}\right)+\left(b^{2}-2 c+\frac{\pi}{3}\right)+\left(c^{2}-2 a\right. \\
\left.+\frac{\pi}{6}\right) \\
=\left(a^{2}-2 a+1\right)+\left(b^{2}-2 b+1\right)+\left(c^{2}-2 c+1\right) \\
+(\pi-3) \\
=(a-1)^{2}+(b-1)^{2}+(c-1)^{2}+\pi-3>0 . \\
\text { If } x \leqslant 0, y \leqslant 0, z \leqslant 0, \text { then } \\
x+y+z \leqslant 0 .
\end{array}
$$
$$
\begin{array}{l}
\text { This contradicts } x+y+z>0 . \\
\text { Therefore, at least one of } x, y, z \text { is greater than } 0 .
\end{array}
$$