How to show that the entropy $H(\operatorname{Pois}(\lambda))$ of a Poisson distribution $\operatorname{Pois}(\lambda)$ is Concave in parameter $\lambda$? i-e
$$f(\lambda)\equiv H(\operatorname{Pois}(\lambda))=-\displaystyle \sum_{x=0}^{\infty} \Big( \frac{e^{-\lambda} \lambda^x}{x!} \Big) \cdot \text{log}_2\Big( \frac{e^{-\lambda} \lambda^x}{x!} \Big) $$ is Concave in $\lambda$?
My Thoughts:
Second derivative Check : It turns out that $\frac{e^{-\lambda} \lambda^x}{x!}$ is log-concave in $\lambda$, since $\frac{d^2}{d \lambda^2}\text{log}_2\Big( \frac{e^{-\lambda} \lambda^x}{x!}\Big)=-\frac{x}{\lambda^2}$. So, each term of the above summation is (at-least) Log-concave because product of log-concave functions result in log-concave function. However, the total summation above is not necessarily log-concave since sum of log-concave functions is not guaranteed to be log-concave (but sum of concave functions is concave, though).
Simulation: Graphical simulations suggest $f(\lambda)$ to be concave in $\lambda$.
Log sum inequality might be helpful, but I don't see a very clear way to use it to prove concavity.
I am assuming base of the logarithm is $e$ instead of $2$ as one can easily convert the base to $e$ by taking a positive constant out.
Note that $f(\lambda)=E_\lambda(\log p_\lambda(X))$ where $p_\lambda$ is the pmf of $Poisson(\lambda)$ distribution and $X\sim Poisson(\lambda)$.
Then using the standard regularity conditions as Poisson belongs to Exponential family, $f''(\lambda)=\dfrac{d^2}{d\lambda^2}E_\lambda(\log p_\lambda(X))=E_\lambda(\dfrac{\partial^2}{\partial \lambda^2}\log p_\lambda(X))=-E_\lambda((\dfrac{\partial}{\partial\lambda}\log p_\lambda(X))^2)\leq0$