Recall the incomplete Gamma function, which is defined as $$ \Gamma(\alpha, x) = \int_x^\infty t^{\alpha - 1} e^{-t} \, dt. $$
My question is whether or not $$ \Gamma(\alpha, x) \leq \frac{3}{2}(x + \alpha)^{\alpha - 1} e^{-x} $$ holds for all $x > 0$ and any $\alpha \geq 0 $.
Below, I prove it holds for all integers $\alpha \geq 0$. Indeed, integration by parts gives $$ \Gamma(\alpha, x) = x^{\alpha - 1} e^{-x} + (\alpha - 1)\Gamma(\alpha- 1, x). $$ In particular, using the fact that $\Gamma(1, x) = \int_x^\infty e^{-t} \, dt = e^{-x}$, it can be verified that for all integers $k \geq 0$, that $$ e^x \Gamma(k + 1, x) = \sum_{j=0}^k x^k (k-j)! \binom{k}{j} \leq \sum_{j=0}^k x^k k^{k-j} \binom{k}{j} = (x + k)^k. $$ Or stated otherwise, for all $x > 0$, $$ \Gamma(\alpha, x) \leq e^{-x} (x + \alpha - 1)^{\alpha - 1} \leq \frac{3}{2} e^{-x} (x + \alpha)^{\alpha - 1} \quad \mbox{for any}~\alpha \in \{1, 2, 3, \dots\}. $$ Moreover, the bound above can also trivially be verified at $\alpha = 0$.
Some thoughts. (A rigorous proof for (1) and (2) is required.)
Let $$F(x) := \frac{3}{2}(x + \alpha)^{\alpha - 1} \mathrm{e}^{-x} - \int_x^\infty t^{\alpha - 1} \mathrm{e}^{-t} \, \mathrm{d} t.$$ We have \begin{align*} F'(x) &= -\frac32(x + \alpha)^{\alpha - 2}\mathrm{e}^{-x}(1 + x) + x^{\alpha - 1} \mathrm{e}^{- x}\\ &= - \frac{\mathrm{e}^{-x} x^{\alpha}(1 + x)}{(x + \alpha)^2}\left( \frac32\left(1 + \frac{\alpha}{x}\right)^{\alpha} - \frac{(x+\alpha)^2}{x(1 + x)}\right). \end{align*}
We split into three cases.
Case 1. $\alpha \ge 1$ and $x > 0$
By Bernoulli inequality, we have $$\frac32\left(1 + \frac{\alpha}{x}\right)^{\alpha} - \frac{(x+\alpha)^2}{x(1 + x)} \ge 1\cdot \left(1 + \frac{\alpha}{x}\cdot \alpha\right) - \frac{(x+\alpha)^2}{x(1 + x)} = \frac{(\alpha - 1)^2}{1 + x}\ge 0.$$ Thus, $F'(x) \le 0$ on $(0, \infty)$. Also, we have $F(\infty) = 0$. Thus, $F(x) \ge 0$ for all $x > 0$.
Case 2. $0 \le \alpha < 1$ and $x > 1$
By Bernoulli inequality, we have $$\left(1 + \frac{\alpha}{x}\right)^{\alpha} = \frac{\left(1 + \frac{\alpha}{x}\right)}{\left(1 + \frac{\alpha}{x}\right)^{1 - \alpha}} \ge \frac{\left(1 + \frac{\alpha}{x}\right)}{1 + \frac{\alpha}{x}\cdot (1 - \alpha)},$$ and $$\frac32\left(1 + \frac{\alpha}{x}\right)^{\alpha} - \frac{(x+\alpha)^2}{x(1 + x)} \ge \frac32 \cdot \frac{\left(1 + \frac{\alpha}{x}\right)}{1 + \frac{\alpha}{x}\cdot (1 - \alpha)} - \frac{(x+\alpha)^2}{x(1 + x)} \ge 0.$$ Thus, $F'(x) \le 0$ on $(1, \infty)$. Also, we have $F(\infty) = 0$. Thus, $F(x) \ge 0$ for all $x > 1$.
Case 3. $0 \le \alpha < 1$ and $0 < x \le 1$
Using $\mathrm{e}^{-t} \ge 1 - t$ and the Bernoulli inequality, we have \begin{align*} \int_x^\infty t^{\alpha - 1} \mathrm{e}^{-t} \, \mathrm{d} t &= \int_0^\infty t^{\alpha - 1} \mathrm{e}^{-t} \, \mathrm{d} t - \int_0^x t^{\alpha - 1} \mathrm{e}^{-t} \, \mathrm{d} t\\ &\le \int_0^\infty t^{\alpha - 1} \mathrm{e}^{-t} \, \mathrm{d} t - \int_0^x t^{\alpha - 1} (1 - t) \, \mathrm{d} t\\ &= \Gamma(\alpha) - \left( \frac{1}{\alpha} x^\alpha - \frac{1}{\alpha + 1}x^{\alpha + 1}\right)\\ &= \Gamma(\alpha) - \frac{x}{x^{1 - \alpha}} \left(\frac{1}{\alpha} - \frac{x}{\alpha + 1}\right)\\ &\le \Gamma(\alpha) - \frac{x}{1 + (x - 1)(1 - \alpha)} \left(\frac{1}{\alpha} - \frac{x}{\alpha + 1}\right). \end{align*}
It suffices to prove that $$f(x) := \frac{3}{2}(x + \alpha)^{\alpha - 1} \mathrm{e}^{-x} - \Gamma(\alpha) + \frac{x}{1 + (x - 1)(1 - \alpha)} \left(\frac{1}{\alpha} - \frac{x}{\alpha + 1}\right) \ge 0.$$
We have \begin{align*} f'(x) &= \frac{(1 - x)(1 + \alpha + x - \alpha x)}{(\alpha + 1)(\alpha + x - \alpha x)^2} - \frac32 (x + \alpha)^{\alpha - 2}\mathrm{e}^{-x}(1 + x)\\ &\le \frac{(1 - x)(1 + \alpha + x - \alpha x)}{(\alpha + 1)(\alpha + x - \alpha x)^2} - \frac32 (x + \alpha)^{\alpha - 2}(1 - x)(1 + x)\\ &= (1 - x)\left(\frac{1 + \alpha + x - \alpha x}{(\alpha + 1)(\alpha + x - \alpha x)^2} - \frac32 (x + \alpha)^{\alpha - 2}(1 + x)\right)\\ &\le 0 \tag{1} \end{align*} where we use $\mathrm{e}^{-x} \ge 1 - x$.
Also, we have $$f(1) = \frac32 (\alpha + 1)^{\alpha - 1}\mathrm{e}^{-1} - \Gamma(\alpha) + \frac{1}{\alpha} - \frac{1}{\alpha + 1} > 0. \tag{2}$$ Thus, $f(x) \ge 0$ for all $0 < x \le 1$.