If $a,b$ are coprime and we already know $\sigma(a)\varphi(a) < a^{2}$ and $\sigma(b)\varphi(b) < b^{2}$, then multiplying these gives $\sigma(ab)\varphi(ab)=\sigma(a)\varphi(a)\sigma(b)\varphi(b) < a^{2}b^{2}=(ab)^{2}$. Thus we just need to check it for $n=p^{k}$ being prime power of a prime $p$. In that case, $\sigma(p^{k})\varphi(p^{k}) = \frac{p^{k+1}-1}{p-1}\cdot (p-1)p^{k-1}= p^{2k}-p^{k-1}< p^{2k}=(p^{k})^{2}$. For the other one, set $c=\frac{1}{\zeta(2)}$, $\zeta(2) : = \sum_{k=1}^\infty \frac{1}{k^{2}}<2$. Let $n=p_{1}^{v_{1}}p_{1}^{v_{1}}...p_{s}^{v_{s}}$. Using multiplicativity and the values at prime powers, we get \[\frac{\sigma(n)\varphi(n)}{n^{2}}=\prod_{i=1}^{s}\frac1{p_{i}^{2v_{i}}}\prod_{i=1}^{s}\frac{p^{v_{i}+1}-1}{p-1}\prod_{i=1}^{s}(p-1)p^{v_{i}-1}= \prod_{i=1}^{s}\left(1-\frac 1{p_{i}^{v_{i}+1}}\right).\] But now \[\left(\prod_{i=1}^{s}\left(1-\frac 1{p_{i}^{v_{i}+1}}\right) \right)^{-1}\leq \left(\prod_{i=1}^{s}\left(1-\frac 1{p_{i}^{2}}\right)\right)^{-1}= \prod_{i=1}^{s}\left(\frac{1}{1-p_{i}^{-2}}\right) \leq \prod_{p \in \mathbb P}\left(\frac{1}{1-p^{-2}}\right) ,\] with $\mathbb{P}$ being the set of primes. But \[\prod_{p \in \mathbb P}\left(\frac{1}{1-p^{-2}}\right) = \prod_{p \in \mathbb P}\sum_{k=0}^\infty (p^{-2})^{k}= \sum_{k=1}^\infty k^{-2}= \zeta (2)\] by factoring (and some closer look at convergences etc.). All together, this gives the result. This also shows that there is no bigger $c$ with that property.