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?



Estimate of a ratio of two incomplete gamma functions
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:


1. In fact we proved estimates from both sides, upper and lower ones;
2. They both are sharp for $x\to y$.


