I faced this problem in my master degree test and unfortunately I can't solve it :(
The problem is:
Lets $(X,Y)$ be a vector of random variables with probability distribution function: $$P(X=x,Y=y)=\int_0^1 \theta^{x+y}(1-\theta)^{2-x-y}\;f(\theta)d\theta\;\Bbb I_A(x) \Bbb I_A(y),\quad A=\{0,1\}$$ Where $f$ is a probability density function such that $\int_0^1f(\theta)d(\theta)=1$. Proof that $$Cov(X,Y)>0$$.
$\mathsf{E}(XY)=1\cdot P(X=1,Y=1)=\int_0^1 \theta^2 f(\theta)d\theta$.
$\mathsf{E}(X)=\mathsf{E}(Y)=1\cdot P(X=1)=P(X=1,Y=0)+P(X=1,Y=1)=\int_0^1 \theta f(\theta)d\theta$.
Since $\text{cov}(X,Y)=\mathsf{E}(XY)-\mathsf{E}(X)\mathsf{E}(Y)$, it suffices to prove $$\int_0^1 \theta^2 f(\theta)d\theta>\left(\int_0^1 \theta f(\theta)d\theta\right)^2$$ Which is a direct consequence of Cauchy-Schwarz, as is shown below: \begin{align}\int_0^1 \theta^2 f(\theta)d\theta & =\left(\int_0^1 \theta^2 f(\theta)d\theta\right)\left(\int_0^1 f(\theta)d\theta\right)\\ &\geq\left(\int_0^1 \theta f(\theta)d\theta\right)^2\end{align} The equality holds iff. $\theta^2 f(\theta)=\alpha f(\theta)$ for some constant $\alpha$, which is impossible.