Note: This post was taken verbatim from this MathStackExchange answer.
Taking cue from reuns'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
$$\prod_{i=1}^{r}{{{p_i}^{\alpha_i}}\cdot\left(1 - {p_i}^{-1} - \mathcal{O}({p_i}^{- 2})\right)} \leq \prod_{i=1}^{r}{{{p_i}^{\alpha_i}}\cdot\left(1 - {p_i}^{-1}\right)} = \varphi(X).$$
Hence, we have the relation
$$D(X) \leq \prod_{i=1}^{r}{\left({p_i}^{\alpha_i} - \sigma({p_i}^{\alpha_i - 1})\right)} \leq \varphi(X),$$
where $D(X)=2X-\sigma(X)$ is the deficiency of $X$, and $\varphi(X)$ is the Euler totient of $X$.
