Estimate of a ratio of two incomplete gamma functions I would like to bound from above the expression
$$
\frac{\Gamma(\alpha,x)-\Gamma(\alpha,y)}{\Gamma(\beta,x)-\Gamma(\beta,y)}
$$
for $x>y>0$.  By plotting the above expression I have found that it should hold
$$
\frac{\Gamma(\alpha,x)-\Gamma(\alpha,y)}{\Gamma(\beta,x)-\Gamma(\beta,y)}< y^{\alpha-\beta}
$$
for $\alpha < \beta$ and
$$
\frac{\Gamma(\alpha,x)-\Gamma(\alpha,y)}{\Gamma(\beta,x)-\Gamma(\beta,y)}< x^{\alpha-\beta}
$$
for $\alpha > \beta$.  To prove the first estimate my guess was to show that a function
$$
f(\alpha)=y^{-\alpha}(\Gamma(\alpha,x)-\Gamma(\alpha,y))
$$
is increasing in $\alpha$.  It seemed natural to me to compute $\partial_\alpha f(\alpha)$ but things started to get rough because $\partial_\alpha \Gamma(\alpha,x)$ has to be expressed as Meijer G-function.  So the question is is there a more elementary proof of the presented estimates?  Are there "better" estimates of which my estimates are special cases?
 A: As @suvrit suggests we have
$$
\frac{\Gamma(\alpha,x)-\Gamma(\alpha,y)}{\Gamma(\beta,x)-\Gamma(\beta,y)}=\frac{\int_y^x t^{\alpha-1}e^{-t}dt}{\int_y^x t^{\beta-1}e^{-t}dt}=\frac{\int_y^x (t^{\alpha-\beta})t^{\beta-1}e^{-t}dt}{\int_y^x t^{\beta-1}e^{-t}dt}.
$$
We can now maximize $t^{\alpha-\beta}$ depending on the sign of $\alpha-\beta$. For any $t\in[y,x]$ we have
$$
t^{\alpha-\beta}\le\begin{cases}x^{\alpha-\beta}\quad &\text{ for } \alpha>\beta\\y^{\alpha-\beta}\quad &\text{ for } \alpha<\beta\\\end{cases}
$$
which gives the requested estimate.
A: In fact it seems to be a consequence of the Cauchy theorem from calculus. Really, by it and a formula for derivative of incomplete gamma function (cf. Wiki for example) we evaluate
$\frac{f(x)-f(y)}{g(x)-g(y)}=\frac{f'(c)}{g'(c)}$
with some intermediate c, $0<y<c<x$ 
and then by the above mentioned formula
$\Gamma^{'}_{x}(s,x)=-\frac{x^{s-1}}{e^x}$
it follows for your expression the exact formula 
$$
\frac{\Gamma(a,x)-\Gamma(a,y)}{\Gamma(b,x)-\Gamma(b,y)}=c^{a-b}
$$
and everything is proved due to $0<y<c<x$.
And even more:


*

*In fact we proved estimates from both sides, upper and lower ones;

*They both are sharp for $x\to y$. 
