# $\Gamma(\alpha+1,t)\geq (t+1)\Gamma(\alpha,t)$:Inequality related to incomplete gamma function

I am trying to prove an inequality which goes as

For $\alpha\geq 1$,$\int_{t}^\infty u^{\alpha}e^{-u}du\geq (t+1) \int_{t}^\infty u^{\alpha-1}e^{-u}du$ for any $t\geq 0$. Using the notation of incomplete gaussian function $\Gamma(\alpha+1,t)\geq (t+1)\Gamma(\alpha,t),\forall t\geq 0$.

For example: We have $$\int_t^\infty u^\alpha e^{-u} du = t^\alpha e^{-t} + \alpha \int_t^\infty u^{\alpha -1} e^{-u} du.$$ The inequality is now equivalent to $$t^\alpha e^{-t} \geq (t+ 1-\alpha) \int_t^\infty u^{\alpha -1} e^{-u} du.$$ If $t\leq \alpha -1$, the inequality is trivial. We consider $t > \alpha -1$. Define the function $$f(t) = \frac{t^\alpha}{t+1 -\alpha} e^{-t} - \int_t^\infty u^{\alpha -1} e^{-u} du,\quad t > \alpha -1.$$ Differentiating $f$, we get $$f'(t) = -\frac{(\alpha -1) t^{\alpha -1}}{(t+ 1-\alpha)^2} e^{-t} < 0.$$ Therefore $$f(t) > \lim_{t\to \infty} f(t) = 0,\quad \forall \alpha -1 < t < \infty.$$
I don't have the rep points to comment, so I've put this in an answer box, but have you tried showing the inequality holds at some convenient $t$ (say $t=0$), then differentiating both sides with respect to $t$ (using product rule on the right). If the derivative on the left is greater than that on the right for all $t$ and $\alpha$ under your conditions, then you've got it.
[update] You can consider this comparing two curves. One is the curve on the left of the inequality, the other on the right. If the curve on the left starts out greater than the one on the right in the range of interest ($t \in [0,\inf)$), and the slope of the curve on the left is always greater than the one on the right, then the function on the left must always be greater. [another update--I wrote this, but I think it's wrong, sorry; I see your point--] [wrong part] This can be true even though the difference between the two curves gets smaller and smaller as $t$ goes to $\inf$. [end wrong part]