It is known that there is no room for improvement to the inequality $\sigma(p)\varphi(p) < p^2$ for primes $p$, since the usual formulae for the sum of divisors and Euler totient functions give $$\sigma(p) = p + 1$$ $$\varphi(p) = p - 1.$$
Here is my question:
Can we improve on the inequality $\sigma(N)\varphi(N) < N^2$ for composite integers $N > 1$?
MY ATTEMPT
Consider the case $N = q^k$ a prime power.
Then we have $$\sigma(N) = \sigma(q^k) = \frac{q^{k+1} - 1}{q - 1}$$ and $$\varphi(N) = \varphi(q^k) = q^{k-1} (q - 1)$$ so that we obtain $$\sigma(N)\varphi(N)=\sigma(q^k)\varphi(q^k)=\bigg(\frac{q^{k+1} - 1}{q - 1}\bigg)\cdot\bigg(q^{k-1} (q - 1)\bigg)=q^{2k} - q^{k-1}.$$
Since both $\sigma$ and $\varphi$ are multiplicative, this means that if we have the canonical prime factorization $$\displaystyle\prod_{i=1}^{w}{{p_i}^{\alpha_i}}$$ for $N$, then we have the following exact expression for $\sigma(N)\varphi(N)$: $$\displaystyle\prod_{i=1}^{w}{\bigg({p_i}^{2\alpha_i} - {p_i}^{\alpha_i - 1}\bigg)},$$ where $w = \omega(N)$ is the number of distinct prime factors of $N$.
Consequently, it may be possible to tweak the upper bound to get a minor(?) or substantial(?) improvement, but I am not seeing it.
UPDATE (1 NOV 2019 - 09:24 AM Manila time)
Basically, I want an upper bound for $\sigma(N)\varphi(N)$, in terms of $N$, that is sharper than $N^2$.
It seems that also for composite $n$ (not even semiprime), the fraction $$\frac{\varphi(n)\sigma(n)}{n^2}$$ can get arbitary close to $\ 1\ $, which would mean that $\ n^2\ $ cannot significantly be improved.
A particular striking number is
$$6110020423763=14173\cdot 17419\cdot 24749$$ giving $$0.99999999009339\cdots $$
So, $\ n^2\ $ seems to be sharp also for composite numbers.