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

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Answer: $\frac{1}{2}$
Analysis: $\max \{|2 x+y|,|x-y|,|1+y|\} \geq \frac{1}{6}(|2 x+y|+2 \cdot|x-y|+3 \cdot|1+y|)$, $\geq \frac{1}{6} \cdot|(2 x+y)-2(x-y)-3(1+y)|=\frac{1}{2}$, which is achieved when and only when $x=0, y=-\frac{1}{2}$.