#### Answer

See proof below.

#### Work Step by Step

$1. \ (\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e})$ Check that the formula holds for $n=-3$
$LHS=2^{-3}=\displaystyle \frac{1}{2^{3}}=\frac{1}{8}$
so, $LHS\displaystyle \geq\frac{1}{8}\quad $ when $n=-3.$
Yes, this is true when $n=-3.$
$2. \ (\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{p})$ Prove that IF the formula holds for any positive integer $n=k\geq-3$, THEN it also holds for the next integer, $n=k+1$.
Let us assume that for $n=k\geq-3$,
$ 2^{k}\geq \displaystyle \frac{1}{8}$
(we call this the step hypothesis)
For $n=k+1$,
the LHS equals$\quad 2^{k+1}$
which we rewrite as
$LHS= 2^{k+1}=(2^{k})\cdot 2\quad $
and we apply the hypothesis
$ 2^{k+1}\displaystyle \geq\frac{1}{8}\cdot 2$
Now, we apply the fact that 2 is greater than 1
$2^{k+1}\displaystyle \geq\frac{1}{8}\cdot 1=\frac{1}{8}$
which makes the formula true for $n=k+1.$
By mathematical induction, the formula is true for any integer $n\geq-3.$