A tough inequality

$(n!)^2 = \displaystyle \prod _{ k=1 }^{ n }{ k(n-k+1) }$

since $n \le k(n-k+1)$ for all $1 \le k \le n$ , we have

$n^n < \displaystyle \prod _{ k=1 }^{ n }{ k(n-k+1) } = (n!)^2$

$\Rightarrow n^{\frac{n}{2}} < n!$

Since both functions are monotonically increasing, we only have to check n = 3. Let y = 3^1.5. thin log(y) = 1.5 log(3) = 1.5 .477121255 = .715681882, and y = 10^.7155681882 = 5.1961522423 < 3! = 6. Ed Gray