Let $k$ represent an integer value. We define function: $f_1(k)=k(1-k t)\frac{\Gamma(L-k,k a_1)}{\Gamma(L-k)}$, for $1\le k\le L-1$, with $kt \le 1$, and $a_1$, $t (<1)$ are some positive constants. $\Gamma(s,x)$ is the upper incomplete Gamma function.
$\bullet$ If we plot $f_1(k)$ we can notice that it starts from $0$, increases to reach a maximum, then decreases to $0$ (at $k= 1/t$).
We assume that $f_1$ reaches its maximum at $k_{m1}$. So $f_1(k_{m1}) > f_1(k^\prime)$, $\forall$ $k^\prime> k_{m1}$.
$\bullet$ Now, we define $f_2(k)=k(1-k t)\frac{\Gamma(L-k,k a_2)}{\Gamma(L-k)}$, where $a_2 >a_1$. Function $f_2$ increases from $0$ to reach its maximum at $k_{m2}$ and then decreases to $0$. So $f_2(k_{m2}) > f_2(k^\prime)$, $\forall$ $k^\prime> k_{m2}$.
Question: My goal is to show that $k_{m1} > k_{m2}$. In other words, I want to show that $f_2(k_{m1}) > f_2(k^\prime)$, $\forall$ $k^\prime> k_{m1}$. Any hint is welcome.
Some properties:
Since $a_2 >a_1$, we have $\Gamma(L-k,k a_1) > \Gamma(L-k,k a_2)$; this is due to the fact that $d\Gamma(s,x)/dx =-x^{s-1} e^{-x} < 0 $, $\forall x$. Thus, we can write $f_1(k) > f_2(k)$.
$\Gamma(L-k_{m1},k_{m1} a_1) > \Gamma(L-k^\prime,k^\prime a_1)$, $\forall$ $k^\prime > k_{m1}$.
$\Gamma(L-k_{m1},k_{m1} a_2) > \Gamma(L-k^\prime,k^\prime a_2)$, $\forall$ $k^\prime > k_{m1}$.
$\Gamma(L-k_{m2},k_{m2} a_2) > \Gamma(L-k^\prime,k^\prime a_2)$, $\forall$ $k^\prime > k_{m2}$.
These result from the following. From Wikipedia, we have $\Gamma(s,x)=(s-1)! e^{-x} \sum_{i=0}^{s-1} x^i /i!$ and $\Gamma(s)=(s-1)!$. So $\Gamma(L-k,k a)= (L-k-1)!e^{-ka} \sum_{i=0}^{L-k-1} (ka)^i /i!$. We know that $\Gamma(s,x)$ decreases with $x$ since the derivative is always $<0$ (proved before). Also, one can remark that $\Gamma(s,x)$ decreases if $s$ decreases. When increasing $k$, $s=L-k$ decreases and $x=ka$ increases. Thus, $\Gamma(L-k,k a)$ decreases with $k$.One can prove that $f_2(k_{m1}) > f_2(k^\prime)$, $\forall$ $k^\prime> k_{m1}$, by showing that $\frac{\Gamma(L-k_{m1},k_{m1} a_2) }{\Gamma(L-k^\prime,k^\prime a_2)} > \frac{\Gamma(L-k_{m1},k_{m1} a_1)}{\Gamma(L-k^\prime,k^\prime a_1)}$.
Example: $L=11$, $t=0.1$, $a_1=1$, $a_2=2$. We get
