Question

Prove $n^n$ $>\mathrm{1.3.5},......(2n-1)$

Answer 1

Consider the first 'n' odd integers 1, 3, 5, ........, (2n - 1)Since all are unequal and positive,
AM [1, 3, 5, ........, (2n - 1)] >
GM [1, 3, 5, ........, (2n - 1)]
$\frac{1+3+5+........+(2n-1)}{n}>[1\times 3\times 5\times ................\times (2n-1){]}^{\frac{1}{n}}$
$\frac{n^2}{n}>[1\times 3\times 5\times ................\times (2n-1){]}^{\frac{1}{n}}$
$n>[1\times 3\times 5\times ................\times (2n-1){]}^{\frac{1}{n}}$
Taking $n^{th}$ power on both sides
$n^n>[1\times 3\times 5\times ................\times (2n-1\left)\right]$