We must show that if $X_{i}$ is a metric space, $\forall i \in \{1, \ldots n\}$ then $\mu(x,y)= \left[ \sum_{i=1}^{n} d_{i}^2(x_{i},y_{i}) \right]^{1/2} $ is a metric for $X= X_{1} \times \ldots \times X_{n}$.
It's clear that $\mu(x,x)=0$, $x \neq y \Rightarrow \mu(x,y)>0$ and that $\mu(x,y)=\mu(y,x)$. The problem starts to show when I try to prove the triangle inequality.
$$\left[ \sum_{i=1}^{n} d_{i}^2(x_{i},z_{i}) \right]^{1/2} \leq \left[ \sum_{i=1}^{n} d_{i}^2(x_{i},y_{i}) \right]^{1/2} + \left[ \sum_{i=1}^{n} d_{i}^2(y_{i},z_{i}) \right]^{1/2} $$
That is equivalent to show that
$$ \sum_{i=1}^{n} d_{i}^2(x_{i},z_{i}) \leq \sum_{i=1}^{n} d_{i}^2(x_{i}y_{i}) + 2\left[ \sum_{i=1}^{n} d_{i}^2(x_{i},y_{i}) \right]^{1/2} \left[ \sum_{i=1}^{n} d_{i}^2(y_{i},z_{i}) \right]^{1/2} + \sum_{i=1}^{n} d_{i}^2(y_{i},z_{i})$$
And that's where I can't continue.
I tried to do this, we know that $d_{i}$ is a métric in $X_{i}$ so $d_{i}(x_{i},z_{i}) \leq d_{i}(x_{i},y_{i}) + d_{i}(y_{i},z_{i})$ which is equivalent to $$d_{i}^2(x_{i},z_{i}) \leq d_{i}^2(x_{i},y_{i}) + 2d_{i}(x_{i},y_{i})d_{i}(y_{i},z_{i}) + d_{i}^2(y_{i},z_{i}) \\ = d_{i}^2(x_{i},y_{i}) + 2\left[d^2{i}(x_{i},y_{i})\right]^{1/2} \left[d_{i}(y_{i},z_{i})\right]^{1/2} + d_{i}^2(y_{i},z_{i})$$.
Aplying sum we get
$$\sum_{i=1}^{n} d_{i}^2(x_{i},z_{i}) \leq \sum_{i=1}^{n} d_{i}^2(x_{i},y_{i}) + 2\sum_{i=1}^{n} \left[ d_{i}^2(x_{i},y_{i})\right]^{1/2} \sum_{i=1}^{n} \left[d_{i}^2(y_{i},z_{i})\right]^{1/2} + \sum_{i=1}^{n} d_{i}^2(y_{i},z_{i})$$
Which is somewhat close but not enought to what we need.