6. If $x, y$ are real numbers, then the minimum value of the maximum number among $|2 x+y|$, $|x-y|$, and $|1+y|$ is

保留源文本的换行和格式，直接输出翻译结果。
6. $\frac{1}{2}$.
$$
\begin{array}{l}
\max \{|2 x+y|,|x-y|,|1+y|\} \\
\geqslant \frac{1}{6}(|2 x+y|+2|x-y|+3|1+y|) \\
\geqslant \frac{1}{6}|(2 x+y)-2(x-y)-3(1+y)| \\
=\frac{1}{2},
\end{array}
$$

The above inequality holds with equality if and only if $x=0, y=-\frac{1}{2}$.