Prove the following

$\frac{1}{9}<\mathrm{I}^{2}<\frac{1}{8}$

$\frac{1}{16}<\mathrm{I}^{2}<\frac{1}{9}$

$\frac{1}{6}<\mathrm{I}^{2}<\frac{1}{2}$

$\frac{1}{8}<\mathrm{I}^{2}<\frac{1}{4}$