Consider three random variables $G,Z,\eta$. $G$ and $Z$ are binary. We assume $$ \begin{aligned} &(1) \quad E(\eta|Z=1)=0,\\ &(2) \quad \Pr(G=1| Z=1)=1. \end{aligned} $$ In turn, $$ \begin{aligned} E(G\eta|Z=1)& =E(G \eta|G=1, Z=1)\Pr(G=1|Z=1)= E(\eta| G=1, Z=1). \end{aligned} $$ Further, observe that $$ \begin{aligned} E(\eta|Z=1)&=E( \eta|G=1, Z=1)\Pr(G=1|Z=1)+E(\eta|G=0, Z=1)\Pr(G=0|Z=1)=E( \eta|G=1, Z=1). \end{aligned} $$ Therefore, $$ (3) \quad E(G\eta|Z=1)=0. $$
Question: Instead of (3), I'd like to write down some sufficient conditions to get the weaker conclusion $$ (3') \quad E(G\eta|Z=1)\geq 0. $$ To this end, I would like to replace (2) with "some" inequality condition, such as $$ \begin{aligned} & E(\eta| Z=1, G=1)\geq E(\eta| Z=1, G=0)\\ & \Pr(G=1|Z=1)\geq \Pr(G=0|Z=1)\\ & \Pr(G=1|Z=1)\geq \Pr(G=1)\\ \end{aligned} $$ or similar. I can keep (1). Any idea?
I am weak in random variables and below is handwaving.
Since the given and the required conditions concerns case $Z=1$, we restrict our consideration to this case. Then Condition (1) becames $E(\eta)=0$ and we need to have $E(G\eta)\ge 0$. But $$E(G\eta)=E(G\eta|G=1)\Pr(G=1)+E(G\eta|G=0)\Pr(G=0)=E(\eta|G=1)\Pr(G=1).$$ So we need to have $E(\eta|G=1)\ge 0$ or $\Pr(G=1)=0$.
But we have $$0=E(\eta)=E(\eta|G=1)\Pr(G=1)+E(\eta|G=0)\Pr(G=0).$$
Thus if the required conclusion fails then $E(\eta|G=1)<0$ and $\Pr(G=1)>0$, and so $E(\eta|G=0)>0$ and $\Pr(G=0)>0$. Thus the proposed condition $E(\eta|G=1)\geq E(\eta| G=0)$ provides the required conclusion.
The other proposed conditions do not provide the required conclusion. Indeed, $\Pr(G=1|Z=1)=\Pr(G=1)$ holds always because we are restricted to the case $Z=1$. To ensure $\Pr(G=1)\geq \Pr(G=0)$ put $\Pr(G=1)=\Pr(G=0)=1/2$. Now put $\eta=1-2G$.