When does the following implication hold:
$$\text{cov}(X,Y)=\mathbb E(XY) \ge0 \Rightarrow \text{cov}(X \log Y)=\mathbb E(X \log Y) \ge0 \tag{1}$$
where $\mathbb E(X)=0$, and $Y$ is a non-negative random variable with $\mathbb E(Y)=1? $
This is equivalent to
$$\text{cov}(Z,Y) \ge0 \Rightarrow \text{cov}(Z,\log Y)\ge0 \tag{2}$$
where both $Z$ and $Y$ are non-negative with $\mathbb E(Z)=\mathbb E(Y)=1.$
I came across with this in my answer, where a condition of form (2) appeared as an equivalent condition for the KL divergence to satisfy the triangle inequality (see O1). I observed that in different cases where $\mathbb E(XY) \ge0$ holds, $\mathbb E(X \log Y) \ge0$ also holds (see O2, O3, and O4, and my final disscussion on their connection with O1).
Positive result:
When the correlation of $X$ and $Y$ is $1$, the above holds. In this case, we have $X=a Y-a$ for some $a\ge 0$, and thus $$\mathbb E(X \log Y)=a \mathbb E(Y\log Y)+a\mathbb E(-\log Y) \ge a(1\log 1-\log 1)=0$$ by Jensen's inequality.
Negative result:
As nicely shown in @RobertIsrae's answer, even if $\text{corr}(X,Y)$ is arbitrarily close to $1$ ($X$ and $Y$ are highly positively correlated), $\text{cov} \left( X, \log Y\right ) \ge 0$ may not hold ($X$ and $\log Y$ can be negatively correlated).
