$(n!)^3<n^n((n+1)/2)^{2n}$
I've been trying to solve this problem. I've tried using these relations: $n^2(n+1/2)^2=$Sum of cubes of first n natural numbers $(n(n+1)/2)=$Sum of first n natural numbers
But still I'm unable to solve this. Can you please help me?
We have for positive integer $n$, $$n!\leq n^n\implies n!^3\leq n^{3n}< (n^3+n^2+n/4)^n=[n (n+1/2)^{2}]^n=n^n(n+1/2)^{2n}$$
Proof by induction: For $n=1$ it holds. Assume that it holds for $n$ we have to prove that it holds for $n+1$, or that: $$ (n+1)^3\lt \frac{(n+1)^{n+1}}{n^n}\frac{(n+1+\frac12)^{2n+2}}{(n+\frac12)^{2n}} $$ which is the same as: $$ (n+1)^2\lt(n+1+\frac12)^2(1+\frac1n)^n(1+\frac1{n+\frac12})^{2n} $$ or that $$ (n+1)^2\lt(n+1+\frac12)^2 $$ which is clearly true.
Proving $$(n!)^3<n^n\left(n+\frac{1}{2}\right)^{2 n}$$ is the same as proving $$3\log(n!) < n\log(n)+2n \log\left(n+\frac{1}{2}\right)$$ Using Stirling approximation for the lhs and Taylor for the rhs $$\text{rhs - lhs}=3 n+\left(1-\frac{3}{2} \log (2 \pi n)\right)-\frac{1}{2 n}+O\left(\frac{1}{n^2}\right)$$ which is true $\forall n \geq 1$.
