Let there be two events (which are disjoint and a partition of the sample space) $G$ and $B$ where $p = Pr(G)$ and $1-p = Pr(B)$. Let $X$ be a random variable and $K$ be a positive constant. Let $D = \mathbb{E}[\min(K, X)|G] - \mathbb{E}[\min(K, X)|B]$ and $E = \alpha \mathbb{E}(X|G) - \alpha \mathbb{E}(X|B)$ where $\alpha = \frac{\mathbb{E}[\min(K,X)]}{\mathbb{E}(X)}$.
Show that $D < E$ if and only if $$\frac{\mathbb{E}(\min(K, X)|G)}{\mathbb{E}(\min(K, X)|B)} < \frac{\mathbb{E}(X|G)}{\mathbb{E}(X|B)} $$
What I've tried so far is to write $\alpha$ as
$$\alpha = \frac{p\mathbb{E}(\min(K, X)|G) + (1-p) \mathbb{E}(\min(K, X)|B)}{p\mathbb{E}(X|G) + (1-p)\mathbb{E}(X|B)} $$ and then I am stuck.
It is only true if we assume something more. Let's $X\sim U(-4,2), K=1$, $G$- $X>=0$, $B$- $X<0$. In this case $D=2.5<4.5=E$, but $$\frac{\mathbb{E}(\min(K, X)|G)}{\mathbb{E}(\min(K, X)|B)}=-\frac{1}{4} >-\frac{1}{2}= \frac{\mathbb{E}(X|G)}{\mathbb{E}(X|B)}. $$
Let's start with $D<E$ \begin{equation} \mathbb{E}(\min(K,X)|G) - \mathbb{E}(\min(K,X)|B)< \frac{\mathbb{E}(\min(K,X))}{\mathbb{E}(X)} \big( \mathbb{E}(X|G) - \mathbb{E}(X|B)\big) \end{equation} Let's assume that $\mathbb{E}(X) > 0$ so we can multiply both sides by $\mathbb{E}(X)$.
Also, as you noticed, $\mathbb{E}(\min(K, X)) = p\mathbb{E}(\min(K, X)|G) + (1-p) \mathbb{E}(\min(K, X)|B)$ and $\mathbb{E}(X)= p\mathbb{E}(X|G) + (1-p)\mathbb{E}(X|B)$. Now we have: \begin{equation} \begin{split} p \mathbb{E}(X|G)\mathbb{E}(\min(K,X)|G)&- &p\mathbb{E}(X|G)\mathbb{E}(\min(K,X)|B) + \\ +(1-p) \mathbb{E}(X|B)\mathbb{E}(\min(K,X)|G)&- (1-p&) \mathbb{E}(X|B)\mathbb{E}(\min(K,X)|B) \\ <\\ p\mathbb{E}(X|G)\mathbb{E}(\min(K, X)|G)&- &p\mathbb{E}(X|B)\mathbb{E}(\min(K, X)|G) + \\ +(1-p)\mathbb{E}(X|G)\mathbb{E}(\min(K, X)|B)&-(1-p&)\mathbb{E}(X|B)\mathbb{E}(\min(K, X)|B) \end{split} \end{equation}
As you can see, we can subtract some components and we are left with: \begin{equation} \begin{split} (1-p) \mathbb{E}(X|B)\mathbb{E}(\min(K,X)&|G)-p\mathbb{E}(X|G)\mathbb{E}(\min(K,X)|B) \\ <\\ (1-p)\mathbb{E}(X|G)\mathbb{E}(\min(K, X)&|B)-p\mathbb{E}(X|B)\mathbb{E}(\min(K,X)|G) \end{split} \end{equation}
Now we add to both sides $p\bigg(\mathbb{E}(X|B)\mathbb{E}(\min(K,X)|G)+\mathbb{E}(X|G)\mathbb{E}(\min(K,X)|B)\bigg)$ to got:
\begin{equation} \mathbb{E}(X|B)\mathbb{E}(\min(K,X)|G) < \mathbb{E}(X|G)\mathbb{E}(\min(K, X)|B) \end{equation}
If we divide by $\mathbb{E}(\min(K, X)|B)\mathbb{E}(X|B)$ we got: $$\frac{\mathbb{E}(\min(K, X)|G)}{\mathbb{E}(\min(K, X)|B)} < \frac{\mathbb{E}(X|G)}{\mathbb{E}(X|B)}. $$
We only have to assure, that both $\mathbb{E}(\min(K, X)|B), \mathbb{E}(X|B)$ have same sign.
If one or three out of $\mathbb{E}(X),\mathbb{E}(\min(K, X)|B), \mathbb{E}(X|B)$ have negative sign, we won't have the dependence. Otherwise it's ok (if $E(X)<0$ you'll have to change the inequality signs respectively after the multiplication, and then after the division).