Verify: If $0<\alpha<1$, $0\leq \beta$, then there exists $M(\alpha,\beta)>0$ so that $$z^\beta e^{-z}\leq Me^{-\alpha z}$$ for all $z\geq 0$
For part (1), I noticed that the left hand side of the inequality is equivalent to a gamma function. So, $$\int_0^\infty z^\beta e^{-z}\,dz=\beta !.$$ If I integrate the right hand side of the inequality I get $$\int_0^\infty M e^{-az}\,dz=M.$$ So if I choose $M(\alpha,\beta)\geq \beta!$. I have proven the inequality?
Is this correct?
EDIT: Proof: Let $f(z)=z^\beta e^{-(1-\alpha)z}$, with $0<\alpha<1$, $0\leq \beta$. Then $$f'(z) = \beta z^{\beta -1}e^{-(1-\alpha)z}+z^\beta e^{-(1-\alpha)z}(\alpha-1) = z^{\beta -1}e^{-(1-\alpha)z}[\beta +z(\alpha-1)] $$ Recall the first derivative test, which states:
"If $f'(x)>0$ on an open interval extending left from $x_0$ and $f'(x)<0$ on an open interval extending right from $x_0$, then $f(x)$ has a local maximum at $x_0$.
Note that $z^{\beta -1}\geq 0$ (since $z\geq 0$), $e^{-(1-\alpha)z}\geq 0$. So $f'(z)=0$ at $z=0$ and $z=-\dfrac{\beta}{\alpha-1}=\dfrac{\beta}{1-\alpha}$, with $z\geq 0$. When $z=0$ we can't test the open interval extending to the left, which doesn't give us any useful information.
When $z=\dfrac{\beta}{1-\alpha}$, choose $z_{-}=\dfrac{\beta-1}{1-\alpha}$ and $z_+=\dfrac{\beta +1}{1-\alpha}$. Then \begin{equation*} \begin{aligned} f'(z_{-}) & = \left(\dfrac{\beta-1}{1-\alpha}\right)^{\beta -1}\exp\left[-(1-\alpha)\dfrac{\beta-1}{1-\alpha}\right]\cdot \left[\beta +\dfrac{\beta-1}{1-\alpha}(\alpha-1)\right] \\ & = \left(\dfrac{\beta-1}{1-\alpha}\right)^{\beta -1}\exp\left[-(\beta-1) \right]\cdot 1 \\ f'(z_{+}) & = \left(\dfrac{\beta+1}{1-\alpha}\right)^{\beta -1}\exp\left[-(1-\alpha)\dfrac{\beta+1}{1-\alpha}\right]\cdot \left[\beta +\dfrac{\beta+1}{1-\alpha}(\alpha-1)\right] \\ & = \left(\dfrac{\beta+1}{1-\alpha}\right)^{\beta -1}\exp\left[-(\beta+1) \right]\cdot -1 \\ \end{aligned} \end{equation*}
However, it isn't obvious which is positive or negative. Help?