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If n ge 1 is a positive integer, then pr...

If n `ge` 1 is a positive integer, then prove that `3^(n) ge 2^(n) + n . 6^((n - 1)/(2))`

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We know that ,
`a^(n) - b^(n) = (a-b) (a^(n-1) + a^(n-2) b + a^(n-3) b^(2) + .. . + b^(n-1))`
`therefore 3^(n) - 2^(n) = 3^(n-1) + 3^(n-2) 2 + 3^(n-3) 2^(2) + … + 2^(n-1)`
Using `A.M. ge G.M` , we get
`(3^(n-1) + 3^(n-2) . 2 + … + 2^(n-1))/(n) ge [(3 * 3^(2) * ... * 3^(n-1)) (2* 2^(2) * ... * 2^(n-1))]^(1//n)`
`3^((n-1)/(2)) * 2((n-1)/(2)) = 6^((n-1)/(2))`
`implies 3^(n) ge 2^(n) + n* 6 ((n-1)/(2))`.
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