I want to prove that :
$\frac{1}{2^r} b^r \geq \frac{1}{2^{3r+3}} b^2$ for all $0<b<1$ and $r>2$
Is it correct to say :
$\frac{1}{2^r} b^r \geq \frac{1}{2^{3r+3}} b^2\Rightarrow 2^{2r+3}\leq b^{2-r}$ as $2-r<0$ then $b^{2-r}>1$ so $2^{2r+3}>1$ which is right ?
