Let $H$ denote the Hellinger distance; i.e., for two discrete distributions $p,q$ (identified with their pmf) over $\mathbb{N}$, $$ H(p,q)^2 = \frac{1}{2}\sum_{n=0}^\infty \left(\sqrt{p(n)}-\sqrt{q(n)}\right)^2 = 1-\sum_{n=0}^\infty \sqrt{p(n)q(n)} $$ and $\operatorname{Poi}(\lambda)$ for the Poisson distribution with parameter $\lambda$.

I would like to know whether there exists a simple proof of the following result:

Let $\lambda>0$, and $\alpha\in[0,1]$. Define $Q= \frac{1}{2}(\operatorname{Poi}((1+\alpha)\lambda) +\operatorname{Poi}((1-\alpha)\lambda) )$. Then $H(\operatorname{Poi}(\lambda),Q) \lesssim \alpha^2\lambda$.

I can easily prove that the *square* Hellinger distance satisfies $H(\operatorname{Poi}(\lambda),Q)^2 \lesssim \alpha^2\lambda$, but the only proof I know for the correct (and tight) statement above is Lemma 4.1 of [VV17]; and I feel there should (?) be a neater and more generalizable argument to establish this seemingly simple bound.

As far as I can tell, the crux of the difficulty (at least in my attempts) is to tightly (upper) bound the quantity $$ \sum_{n=0}^\infty \frac{\lambda^n}{n!}\sqrt{\frac{e^{\lambda\alpha}(1-\alpha)^n+e^{-\lambda\alpha}(1+\alpha)^n}{2}} $$ where the obvious attempts (e.g., AM-GM) only yield the loose bound mentioned above.

[VV17] Valiant, Gregory; Valiant, Paul. *An automatic inequality prover and instance optimal identity testing.* SIAM J. Comput. 46 (2017), no. 1, 429--455.
https://theory.stanford.edu/~valiant/papers/instanceOptFull.pdf