设 ${{x}_{1}},{{x}_{2}},\cdots ,{{x}_{7}}\geqslant 0$ 且 ${{x}_{1}}+{{x}_{2}}+\cdots +{{x}_{7}}=1$，
【难度】
【出处】
无
【标注】
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求 $\min \left\{ {{x}_{1}}+{{x}_{2}}+{{x}_{3}},{{x}_{2}}+{{x}_{3}}+{{x}_{4}},\cdots ,{{x}_{5}}+{{x}_{6}}+{{x}_{7}} \right\}$ 的最大值．标注答案略解析$1=\left({{x}_{1}}+{{x}_{2}}+{{x}_{3}} \right)+{{x}_{4}}+\left({{x}_{5}}+{{x}_{6}}+{{x}_{7}} \right)\geqslant \min +0\text{+}\min =2\min $
故 $\min \leqslant \dfrac{1}{2}$
${{x}_{1}}=0$，${{x}_{2}}=0$，${{x}_{3}}=\dfrac{1}{2}$，${{x}_{4}}=0$，${{x}_{5}}=\dfrac{1}{2}$，${{x}_{6}}=0$，${{x}_{7}}=0$ 时符合 -
求 $\max \left\{ {{x}_{1}}+{{x}_{2}}+{{x}_{3}},{{x}_{2}}+{{x}_{3}}+{{x}_{4}},\cdots ,{{x}_{5}}+{{x}_{6}}+{{x}_{7}} \right\}$ 的最小值．标注答案略解析$3=\left({{x}_{1}}+{{x}_{2}}+{{x}_{3}} \right)+{{x}_{4}}+\left({{x}_{5}}+{{x}_{6}}+{{x}_{7}} \right)$
$+\left({{x}_{1}}+{{x}_{2}}+{{x}_{3}} \right)+\left( {{x}_{4}}+{{x}_{5}}+{{x}_{6}}\right)+{{x}_{7}}$
$+{{x}_{1}}+\left({{x}_{2}}+{{x}_{3}}+{{x}_{4}} \right)+\left( {{x}_{5}}+{{x}_{6}}+{{x}_{7}}\right)$
$\leqslant 6\max +\left({{x}_{1}}+{{x}_{4}}+{{x}_{7}} \right)\leqslant 6\max +1$
故 $\max \geqslant \dfrac{1}{3}$
${{x}_{1}}=\dfrac{1}{3}$，${{x}_{2}}=0$，${{x}_{3}}=0$，${{x}_{4}}=\dfrac{1}{3}$，${{x}_{5}}=0$，${{x}_{6}}=0$，${{x}_{7}}=\dfrac{1}{3}$ 时符合．
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