First of all, your $k$ and $m$ are the same. To make things clearer, I use the notation $k$ instead of $m$. By induction, we assume that
$$\left(\sum\limits_{n=1}^k n\right)\cdot\left(\sum\limits_{n=1}^k \frac1n\right) \geq k^2,$$
and we're going to show that
$$\left(\sum\limits_{n=1}^{k+1} n\right)\cdot\left(\sum\limits_{n=1}^{k+1} \frac1n\right) \geq (k+1)^2.\tag{$*$}$$
You have shown that
\begin{align*}
\left(\sum\limits_{n=1}^{k+1} n\right)\cdot\left(\sum\limits_{n=1}^{k+1} \frac1n\right)&\geq k^2+(k+1)\sum_{n=1}^k\frac1n+\frac1{k+1}\sum_{n=1}^kn+1\\
&=k^2+\frac k2+1+(k+1)\sum_{n=1}^k\frac1n.
\end{align*}
As a result, to show $(*)$, it suffices to prove that
$$(k+1)\sum_{n=1}^k\frac1n\geq\frac{3k}{2},$$
which is equivalent to
$$\sum_{n=1}^k\frac1n\geq\frac{3k}{2(k+1)}.\tag{$**$}$$
Now, we can use another induction to show $(**)$. Clearly $(**)$ holds for $k=1$. Assume that $(**)$ holds for $k$, we need to prove
$$\sum_{n=1}^{k+1}\frac1n\geq\frac{3(k+1)}{2(k+2)}.\tag{1}$$
Indeed, we have
$$\sum_{n=1}^{k+1}\frac1n=\sum_{n=1}^{k}\frac1n+\frac1{k+1}\geq \frac{3k}{2(k+1)}+\frac1{k+1}=\frac{3k+2}{2(k+1)}.$$
So, it suffices to show that $\frac{3k+2}{2(k+1)}\geq\frac{3(k+1)}{2(k+2)}$, which is equivalent to $(3k+2)(k+2)\geq3(k+1)^2$. We have simply
$$(3k+2)(k+2)\geq3(k+1)^2=2k+1>0.$$
Therefore $(1)$ holds and hence $(**)$ holds for all $k$. This completes the proof.
Quicker proof. Your desired inequality is a direct consequence of Cauchy–Schwarz inequality.
Quicker proof of $(**)$ from comments: $(**)$ holds clearly for $k=1$. And for $k\geq 2$, note that
$$\sum_{n=1}^k\frac1n\geq1+\frac12=\frac32=\frac{3(k+1)}{2(k+1)}>\frac{3k}{2(k+1)}.$$
