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A Harder Inequality Proof

Prove that:

\(\frac { 1 }{ 1+{ r }_{ 1 } } +...+\frac { 1 }{ 1+{ r }_{ n } } \ge \frac { n }{ 1+\sqrt [ n ]{ { r }_{ 1 }...{ r }_{ n } } } .\)

Note by Carlos E. C. do Nascimento
3 years ago

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