Prove that $(x+1)(2n+1)>= 6(n!)^(2/n)$ ?

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Answer 1 · 2017-06-22T12:32:21

See below.

Using the Stirling's asymptotic approximation formula

$n! approx sqrt(2pi n)(n/e)^n$ we have

$6(n!)^(2/n) approx 6(sqrt(2 pi n)(n/e)^n)^(2/n) = 6(2pi n)^(1/n)(n/e)^2$ then

$(x+1)(2n+1) ge 6(2pi n)^(1/n)(n/e)^2$ implies that $x ge n$

So the affirmation is true only for $x ge n$

Prove that (x+1)(2n+1)>= 6(n!)^(2/n)(x+1)(2n+1)>= 6(n!)^(2/n)?