What is a sharp upper bound for $$\prod_{i=1}^{r}{\left({p_i}^{\alpha_i} - \sigma({p_i}^{\alpha_i - 1})\right)}?$$
Here, the $p_i$'s are primes, and $\sigma = \sigma_1$ is the classical sum-of-divisors function. You may assume that $$X = \prod_{i=1}^{r}{{p_i}^{\alpha_i}},$$ and that $r = \omega(X)$ is the number of distinct prime factors of $X$.
(Note: This question is related to this earlier MSE post.)
Taking cue from user1952009's comments, we have $$\prod_{i=1}^{r}{\left({p_i}^{\alpha_i} - \sigma({p_i}^{\alpha_i - 1})\right)} = \prod_{i=1}^{r}{\left({p_i}^{\alpha_i} - \frac{{p_i}^{\alpha_i} - 1}{p_i - 1}\right)}.$$
Since $$\frac{p^{\alpha} - 1}{p - 1} = p^{\alpha - 1} + \mathcal{O}(p^{\alpha - 2})$$ we obtain $$\prod_{i=1}^{r}{\left({p_i}^{\alpha_i} - \frac{{p_i}^{\alpha_i} - 1}{p_i - 1}\right)} = \prod_{i=1}^{r}{\left({p_i}^{\alpha_i} - {p_i}^{\alpha_i - 1} - \mathcal{O}({p_i}^{\alpha_i - 2})\right)} = \prod_{i=1}^{r}{{{p_i}^{\alpha_i}}\cdot\left(1 - {p_i}^{-1} - \mathcal{O}({p_i}^{- 2})\right)}.$$
This last quantity is bounded by $$\prod_{i=1}^{r}{{{p_i}^{\alpha_i}}\cdot\left(1 - {p_i}^{-1} - \mathcal{O}({p_i}^{- 2})\right)} \leq \prod_{i=1}^{r}{{{p_i}^{\alpha_i}}\cdot\left(1 - {p_i}^{-1}\right)} = \phi(X).$$
Hence, we have the relation $$D(X) \leq \prod_{i=1}^{r}{\left({p_i}^{\alpha_i} - \sigma({p_i}^{\alpha_i - 1})\right)} \leq \phi(X),$$ where $D(X)=2X-\sigma(X)$ is the deficiency of $X$.