Metric on Cartesian Product I'm currently stuck on an exercise where I need to show that if $(M,d)$ and $(N,\rho)$ are metric spaces, then $d_2((a,x),(b,y)) := \sqrt{d(a,b)^2 + \rho(x,y)^2}$ defines a metric on $M\times N$. I've already shown that the basic criteria of a metric are met, but I still need to show that the triangle ineqaulity holds.
I've tried using the triangle inequality on $d$ and $\rho$ and estimating the square root with $\sqrt{a^2+b^2}\leq a+b$, but regardless of which order I've used them in, I seem to overestimate my target. I thought maybe I somehow need to use Cauchy-Schwarz, but I don't know how.
 A: Let $(a,b,c) \in M$ and $(x,y,z) \in N$. You want to prove that
$$\sqrt{d(a,c)^2 + \rho(x,z)^2} \leq \sqrt{d(a,b)^2 + \rho(x,z)^2} + \sqrt{d(b,c)^2 + \rho(y,z)^2}$$
You have, by triangle inequality for $d$ and $\rho$ :
$$d(a,c)^2 + \rho(x,z)^2 \leq d(a,b)^2 + d(b,c)^2 + 2 d(a,b)d(b,c) + \rho(x,y)^2 + \rho(y,z)^2 + 2 \rho(x,y)\rho(y,z) \quad (1)$$
But on the other hand, you know that
$$(\rho(x,z)d(b,c) - \rho(y,z)d(a,b))^2 \geq 0$$
so
$$\rho(x,z)^2 d(b,c)^2 + \rho(y,z)^2d(a,b)^2 \geq 2 \rho(x,z)d(b,c)\rho(y,z)d(a,b)$$
Adding $d(a,b)^2d(b,c)^2 + \rho(x,y)^2 \rho(y,z)^2$ to this relation, you get
$$d(a,b)^2d(b,c)^2 + \rho(x,y)^2 \rho(y,z)^2 +\rho(x,z)^2 d(b,c)^2 + \rho(y,z)^2d(a,b)^2 \geq d(a,b)^2d(b,c)^2 + \rho(x,y)^2 \rho(y,z)^2 + 2 \rho(x,z)d(b,c)\rho(y,z)d(a,b)$$
i.e.
$$(d(a,b)^2 + \rho(x,z)^2)(d(b,c)^2 + \rho(y,z)^2) \geq (d(a,b)d(c,d)+\rho(x,y)\rho(y,z))^2$$
so
$$2 \sqrt{(d(a,b)^2 + \rho(x,z)^2)(d(b,c)^2 + \rho(y,z)^2)} \geq 2(d(a,b)d(c,d)+\rho(x,y)\rho(y,z)) $$
Injecting this in the equation $(1)$, you get
$$d(a,c)^2 + \rho(x,z)^2 \leq d(a,b)^2 + d(b,c)^2 + \rho(x,y)^2 + \rho(y,z)^2 + 2\sqrt{(d(a,b)^2 + \rho(x,z)^2)(d(b,c)^2 + \rho(y,z)^2)}$$
which is equivalent to
$$d(a,c)^2 + \rho(x,z)^2 \leq \left(\sqrt{d(a,b)^2+\rho(x,z)^2}+\sqrt{d(b,c)^2 + \rho(y,z)^2}\right)^2$$
i.e. 
$$\sqrt{d(a,c)^2 + \rho(x,z)^2} \leq \sqrt{d(a,b)^2 + \rho(x,z)^2} + \sqrt{d(b,c)^2 + \rho(y,z)^2}$$
A: Let $m,m',m''\in M$ and $n,n',n''\in N .$ Let $p=d(m,m'), q=d(m',m''),r=d(m,m'').$ Let $s=\rho(n,n'), t=\rho (n',n''), u=\rho (n,n'').$
Let $A= (p^2+s^2)^{1/2}$ and $B=(q^2+t^2)^{1/2}.$ We wish to prove that $A+B\geq (r^2+u^2)^{1/2}.$
Let $C= ((p+q)^2+(s+t)^2)^{1/2}.$ Since $d$ and  $\rho$ are metrics, we have $p+q\geq r\ge 0$ and $s+t\geq u\ge 0,$ so $C\ge (r^2+u^2)^{1/2}.$ So it suffices to prove that $A+B\geq C.$
Since $A,B,C \geq  0$ we have $$ A+B\geq C \iff (A+B)^2\geq C^2 \iff$$ $$\iff p^2+s^2+q^2+t^2+2 ((p^2+s^2)(q^2+t^2))^{1/2}\geq p^2+q^2+s^2+t^2+2pq+2st$$ $$\iff ((p^2+s^2)(q^2+t^2))^{1/2}\geq pq+st \iff$$ $$\iff (p^2+s^2)(q^2+t^2)\geq (pq+st)^2\iff$$ $$\iff (pt-qs)^2\geq 0.$$
In the last two lines above, we have used the identity $(p^2+s^2)(q^2+t^2)=(pq+st)^2+(pt-qs)^2.$ 
