Question

Show that ${\left(|\underset{–}{n}\right)}^3>n^n{\left(\frac{n+1}{2}\right)}^{2n}$.

Answer 1

${\underset{–}{|n}}^3<n^n{\left(\frac{n+1}{2}\right)}^{2n}$

According to AM-GM inequality, where arithematic mean is always $\ge$ geometric mean.

So,

$\frac{1^3+2^3+3^3...n^3}{n}\ge {(1^3\times 2^3...\times n^3)}^{\frac{1}{n}}$

Sum of cube of $n$ natural number will be ${\left(\frac{n(n+1)}{2}\right)}^2$ and $1\times 2\times 3\times ...\times n=\underset{–}{|n}$

Therefore,

$\frac{{\left(\frac{n(n+1)}{2}\right)}^2}{n}\ge {\underset{–}{|n}}^{\frac{3}{n}}$

${\underset{–}{|n}}^3\le \frac{{\left(\frac{n(n+1)}{2}\right)}^{2n}}{n^n}$

${\underset{–}{|n}}^3\le n^n{\left(\frac{n+1}{2}\right)}^{2n}$