Show that $d(x,y)$ in a metric on $X$. Let $d_a(x,y)=7|x-y|$ and $d_b(x,y)=|x+y|$ be metrics on set $X$. Show that $d(x,y)=d_a(x,y) + d_b(x,y)$ is also a metric on $X$.
Would I be correct in writing $d_a(x,y) + d_b(x,y)$ as $7|x-y| + |x+y|$? And then applying the regular properties of a metric? If this is the case then how would we prove the triangle inequality?
 A: This is NOT even a metric, in fact even $d_b(x,y)$ is NOT a metric.
$$0=d_b(x,y) := |x+y| \iff x=-y$$ 
A: In response to the previous comment in my other answer, I will show that the sum of two metrics $(d_1(x,y)$ and $d_2(x,y))$ defines a metric ($d(x,y) = d_1(x,y)+d_2(x,y))$
Proof:
Suppose $d(x,y)=0$ then
$$d(x,y) = d_1(x,y) + d_2(x,y) = 0$$
As $d_1$ and $d_2$ are metrics, they must satisfy $d_1(x,y),d_2(x,y)\ge 0$ for all $x,y \in X$.  But this means that they must be precisely zero, but metrics only equal zero when their arguments are equal, hence $x=y$.
Conversely, suppose $x=y$, then $d_1(x,y)=d_2(x,y)=0$ because these are metrics, hence $d(x,y)=0$.
Symmetry,
$$d(x,y) = d_1(x,y) + d_2(x,y) = d_2(y,x)+d_1(y,x)=d(y,x)$$
where the middle equals sign is true because $d_1, d_2$ are metrics and hence symmetric in their own right.
Triangle Inequality
$$d(x,z) = d_1(x,z) + d_2(x,z) \le [d_1(x,y) + d_1(y,z)] + [d_2(x,y) + d_2(y,z)]=$$ $$=[d_1(x,y) + d_2(x,y)] + [d_1(y,z) + d_2(y,z)]= d(x,y) + d(y,z) $$
where the $\le$ is true because $d_1,d_2$ are metrics and hence satisfy the triangle inequality on their own.
