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Originally the problem is to prove that $n! \geq n^{n/2}$.

I reduced this to: $n! \geq (\sqrt{n})^n$ so that:

Prove that $\frac{n!}{(\sqrt{n})^n} \geq 1$.

Each term in $n!$ is divided by the $\sqrt{n}$ and the multiplication should leave it $\geq 1$.

Some advice.

  • 1
    Essentially duplicate of http://math.stackexchange.com/questions/46892/how-do-you-prove-that-nn-is-on2 and http://math.stackexchange.com/questions/136626/lim-limits-n-to-infty-sqrtnn-is-infinite2012-10-09
  • 0
    Given the fact that ab <= a + b - 1.2012-10-09
  • 0
    Given the fact that (after proof) ab <= a + b - 1. If we write n!^2 = (n * 1)*(n-1)*2*(n-2)*3*........2*(n-1)*(1*n). n*1>=n+1-1<=n, (n-1)*2 >= n - 1 + 2 -1<=n,... According to the theorem each term >= n. So that n!^22012-10-09

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