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
$\begingroup$
$\endgroup$
2
-
$\begingroup$ Hint: $1^3+2^3+...+n^3=\frac{n(n+1)(2n+1)}{6}$ $\endgroup$– Peter ForemanFeb 2, 2019 at 13:55
-
$\begingroup$ That is the sum of first natural squares I think not of cubes $\endgroup$– ThomasFeb 2, 2019 at 16:42
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
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.
answered Feb 2, 2019 at 14:34
$\begingroup$
$\endgroup$
You can also use Stirling approximation and Taylor expansions to show that $$a_n=n \log (n)+2 n \log \left(\frac{n+1}{2}\right) - 3 \log(n!) >0$$ This would give $$a_n=\left(2-\frac{3}{2} \log \left({2 \pi }\right)\right)+ (3-\log (4))\,n-\frac{3}{2} \log \left({n}\right)-\frac{5}{4 n}+O\left(\frac{1}{n^2}\right)$$ which is positive $\forall\, n \geq 2$.
answered Feb 2, 2019 at 16:13