Solution:

We note that equality holds for $n=1$. We prove the inequality when $n>1$ and show that it is strict in this case.

By the AM-GM inequality, $\frac{1}{\varphi(n)}+\frac{1}{\sigma(n)} \geq \frac{2}{\sqrt{\varphi(n) \sigma(n)}}$. Hence, we need only show that $\varphi(n) \sigma(n)<n^{2}$, or equivalently, $\frac{\varphi(n) \sigma(n)}{n^{2}}<1$. We note that $f(n):=\frac{\varphi(n) \sigma(n)}{n^{2}}$ is multiplicative, i.e., $f(m n)=f(m) f(n)$ whenever $\operatorname{gcd}(m, n)=1$, and so it suffices to show that $f(n)<1$ when $n=p^{k}$ for some prime $p$ and some integer $k \geq 1$. However,
$$\begin{aligned} f\left(p^{k}\right) & =\frac{\varphi\left(p^{k}\right) \sigma\left(p^{k}\right)}{\left(p^{k}\right)^{2}}=\frac{\varphi\left(p^{k}\right)}{p^{k}} \cdot \frac{\sigma\left(p^{k}\right)}{p^{k}}=\frac{(p-1) p^{k-1}}{p^{k}} \cdot \frac{1}{p^{k}} \sum_{i=0}^{k} p^{i} \\ & =\frac{p-1}{p} \cdot \sum_{i=0}^{k} \frac{1}{p^{k-i}}=\frac{p-1}{p} \cdot \sum_{i=0}^{k} \frac{1}{p^{i}}=\frac{p-1}{p} \cdot \frac{1-\frac{1}{p^{k+1}}}{1-\frac{1}{p}} \\ & <\frac{p-1}{p} \cdot \frac{1}{1-\frac{1}{p}}=1 \end{aligned}$$
This completes the proof.