Prove the inequality without using the concept of Arithmetic and Geometric mean inequality

Question

$\displaystyle ( 2m+1) r^{m}( 1-r) < 1-r^{2m+1}$ where $r<1$ and m is positive integer. I can prove it by concept of arithmetic and geometric mean inequality. But I am curious to know whether there is any other method to prove it as in the book nothing has been mentioned about arithmetic-geometric mean inequality throughout the chapter Geometric progression (Higher Algebra by Hall and Knight).Can this be solved only by theorem of summation of series etc.

Answer 1

Assuming $0<r<1$, $$\frac{1-r^{2m+1}}{r^m(1-r)}=\frac{1+r+r^2+\cdots+r^{2m}}{r^m} =\frac1{r^m}+\frac1{r^{m-1}}+\cdots+\frac1r+1+r+r^2+\cdots+r^m.$$ If you can prove that $1/r^k+r^k>2$ for $k=1,2,\ldots,m$ then this will be $>2m+1$.