It has been suggested that the inequality be reversed. Certainly we get a correct inequality, unfortunately a fairly uninteresting one. We propose that instead we let $f(n)=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2^n},$ and show that $f(n)\ge 1+\dfrac{n}{2}$ for every integer $n \ge 0$.
It is clear that the result holds when $n=0$. Suppose that the inequality holds at $n=k$. We show that it holds at $n=k+1$. We have $f(k+1)=f(k)+\frac{1}{2^k+1}+\frac{1}{2^k+2}+\cdots +\frac{1}{2^{k+1}}.\tag{$1$}$ Each of $\frac{1}{2^k+1}$, $\frac{1}{2^k+2}$, and so on up to $\frac{1}{2^{k+1}}$ is $\ge \frac{1}{2^{k+1}}$, and there are $2^k$ such terms. So their sum is $\ge \frac{1}{2}$. Thus $f(k+1)\ge f(k)+\frac{1}{2}\ge 1+\frac{k}{2}+\frac{1}{2}=1+\frac{k+1}{2},$ and now we have done the induction step.
Remark: This is a somewhat formalized version of the usual argument that the harmonic series diverges.