Prove that
$$n^n\left(\frac{n+1}{2}\right)^{2n}\geq (n!)^3$$ for a natural number $n$.
what i try
i have use AM GM inequality
$$\frac{1^3+2^3+3^3+\cdots +n^3}{n}\geq ((n!))^{\frac{1}{3n}}$$
how i prove question inequality help me please
Prove that $n^n\left(\frac{n+1}{2}\right)^{2n}\geq (n!)^3$ for natural number $n$
inequality
Best Answer
First of all, by expanding the terms we obtain$$n^n\bigg(\frac{n+1}{2}\bigg)^{2n}{={1\over n^n}\cdot \bigg(\frac{n(n+1)}{2}\bigg)^{2n}\\={1\over n^n}\cdot (1+2^3+\cdots +n^3)^n.}$$By substitution and taking the $n$-th root, we need to prove that $${1+2^3+\cdots+n^3\over n}\ge \sqrt[n]{n!^3}=\sqrt[n]{1^3\cdot 2^3 \cdots n^3},$$which is straight-forward by AM-GM.