Squared Hellinger Distance subadditive for Product measures How can I show that the squared Hellinger Distance is subadditive for Product measures?
We have $\mathbb{P} = \otimes_{i=1}^n \mathbb{P_i}$ and  $\mathbb{Q} = \otimes_{i=1}^n \mathbb{Q_i}$ probability measures on $(\mathcal{X}, \mathcal{F})$, both being dominated by a $\sigma$-finite measure $\mu$.
The corresponding $\mu$-densities will be called $p, p_i, q, q_i$.
The squared Hellinger distance is defined as
$H^2(\mathbb{P}, \mathbb{Q}) = \int_\mathcal{X} \left( \sqrt{p(x)} - \sqrt{q(x)} \right)^2 \mu(dx)$
I want to show: 
$H^2(\mathbb{P}, \mathbb{Q}) \leq \sum_{i=1}^n H^2(\mathbb{P}_i, \mathbb{Q}_i) $
With
$H^2(\mathbb{P}, \mathbb{Q}) = \int_\mathcal{X} \left( \sqrt{ \prod_{i=1}^n p_i(x_i)} - \sqrt{\prod_{i=1}^n q_i(x_i)} \right)^2 \mu(dx) = 
2 -2\int_\mathcal{X} \sqrt{\prod_{i=1}^n p_i(x)q_i(x)} \mu(dx) $ 
 A: It suffices to consider the case $n=2$; the general case follows by iteration.
First of all, note that we can write $\mu = \mu_1 \otimes \mu_2$ where $\mu_i := \mu \circ \pi_i$ denotes the image measure of the projection $\pi_i: \mathbb{R}^2 \to \mathbb{R}, (x_1,x_2) \mapsto x_i$ with respect to $\mu$. Then, by Tonelli's theorem,
$$\begin{align*} 1- 2 H^2(\mathbb{P},\mathbb{Q}) &= \int \sqrt{pq} \, d\mu \\ &= \int \sqrt{p_1 q_1} \, d\mu_1 \cdot \int \sqrt{p_2 q_2} \, d\mu_2 \\ &= \left(1- \frac{1}{2} H^2(\mathbb{P}_1,\mathbb{Q}_1) \right) \left( 1- \frac{1}{2} H^2(\mathbb{P}_2,\mathbb{Q}_2) \right). \end{align*}$$
Consequently,
$$\begin{align*} H^2(\mathbb{P},\mathbb{Q}) &= 2 - 2 \left(1- \frac{1}{2} H^2(\mathbb{P}_1,\mathbb{Q}_1) \right) \left( 1- \frac{1}{2} H^2(\mathbb{P}_2,\mathbb{Q}_2) \right) \\ &= H^2(\mathbb{P}_1,\mathbb{Q}_1) + H^2(\mathbb{P}_2,\mathbb{Q}_2) - \frac{1}{2} H^2(\mathbb{P}_1,\mathbb{Q}_1) H^2(\mathbb{P}_2,\mathbb{Q}_2) \\ &\leq H^2(\mathbb{P}_1,\mathbb{Q}_1) + H^2(\mathbb{P}_2,\mathbb{Q}_2). \end{align*}$$
