Given a pair $(X,Y)$ of r.v.s such that: $$X \sim \text{Poisson}(\lambda)\quad \text{and}\quad Y \sim B(\frac{\lambda}{1+\lambda})$$ with $X,Y$ independent, determine a one-dimensional sufficient statistic for $\lambda$ and the UMVUE (uniformly minimum variance unbiased estimator) for $\lambda$.
Attempt:
$$P(X,Y\mid(x_1,x_2,\dots, x_n, y_1,y_2,\dots y_n)) = P(X\mid(x_1,x_2, \dots x_n))\times P(Y|(y_1,y_2, \dots y_n)) = \prod_{i=1}^n\frac{\lambda^{x_i}e^{-\lambda}}{x_i!} \times \frac{\lambda^{y_i}}{1+\lambda} = \frac{\lambda^{\sum (x_i+y_i)} e^{-n\lambda}}{(1+\lambda)^n \prod x_i!} $$
Using the factorisation theorem, $\sum(x_i+y_i)$ is a sufficient statistic. It is however not an unbiased estimator of $\lambda$ because:
$$E\left(\sum(x_i+y_i)\right) = n\left(\lambda + \frac{\lambda}{1+\lambda}\right)$$
A trivial unbiased estimator would be $\frac{\sum x_i}{n}$, but then it is not sufficient.
To utilise Rao-Blackwell Theorem, $\sum x_i$ should be conditioned on $\sum x_i + y_i$. I am not able to proceed further.