Question

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

Answer 1

See below.

Explanation:

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`