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show that the AM>=GM
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AM is arithmetric mean, GM is geometric mean
\[AM \ge GM\]
\[x _{1},x _{2},x _{3},...x _{n}\] \[AM=\frac{ x _{1}+...+x _{n} }{ n }\]
\[GM=\sqrt[n]{x _{1}\times \times \times x _{n}}\]
eg \[2,4,\] \[AM=3,GM=\sqrt[]{8}\] AM>GM
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