Let $X'$ and $X$ be metric spaces with measures $d$ and $d'$. Show that $m((p,p'), (q,q')) = \sqrt{d(p, q)^2 + d'(p,q)^2}$ is a metric. For the given metric  $m((p,p'), (q,q')) = \sqrt{d(p, q)^2 + d'(p,q)^2}$. I have proven the first two properties for a metric hold (this is rather trivial). I am unable to prove the triangle inequality. My work so far is as follows:
$$m((p,p'), (q,q')) = \sqrt{d(p, q)^2 + d'(p',q')^2} \le \sqrt{(d(p,r) + d(r, q))^2 + (d'(p',r') + d'(r', q'))^2} \\ \le \sqrt{(d(p,r) + d(r, q))^2} + \sqrt{(d'(p',r') + d'(r', q'))^2}$$
How can I continue from here? Was invoking $\sqrt{x+y} \le \sqrt x + \sqrt y$ helpful?

Hint:
Fixing notation, I assume you mean
$$m((p,p'), (q,q')) = \sqrt{ d(p,q)^2 + d'(p',q')^2}$$
This is a generalization of a Euclidean norm on $\mathbb R^2$. You can prove the triangle inequality the same way as we do there. Namely, write
$$m((p,p'), (q,q'))^2 = \langle (d(p,q),d'(p',q')), (d(p,q),d'(p',q')) \rangle_{\mathbb R^2}^2$$
Your proof will require the Cauchy-Schwarz inequality.
Try writing out the proof for $\mathbb R^2$. Then write it out for your metric $m$.
