how to prove this is a metric given the following conditions I need help wrapping my head around the concepts of metrics and how to prove that something is a metric. For example, prove that if $p_1$ and $p_2$ are metrics in $X$, then $p_1 + p_2$ and $\max\{p_1, p_2\}$ are also metrics. Are the functions $\min \{p_1, p_2 \}$ and $p_1*p_2$ metrics and why? 
 A: The problematic part is the triangle inequality.  For example, if $p_1(a,b) = 1$, $p_1(b,c) = 2$, $p_1(a,c) = 3$ while $p_2(a,b) = 2$, $p_2(b,c) = 1$, $p_2(a,c) = 3$, do you see why $\min(p_1,p_2)$ and $p_1 p_2$ don't work?
A: All you have to do is to check the definition:
You know that a metric $d$ is a function $d \colon X \to [0,\infty]$ such that the following three properties are satisfied:
$1)$ $d(x,y) = 0$ if and only if $x = y$
$2)$ $d(x,y) = d(y,x)$ for any $x,y \in X$
&$3)$ $d(x,y) \le d(x,z) + d(z,y)$ for any $x,y,z \in X$.
I will show that if $d_1$ and $d_2$ are metric on the same set $X$ then $d_1 + d_2$ is  a metric. You can play with all the others :D
By definition $(d_1 + d_2)(x,y) = 0$ if and only if $d_1(x,y) + d_2(x,y) = 0$ since they are positive this happen if and only if $d_1(x,y) = 0$ and $d_2(x,y) = 0$. But $d_1$ is a metric then this can happen of and only if $x = y$.
Symmetry is even simpler if possible: $(d_1 + d_2)(x,y) = d_1(x,y) + d_2(x,y) = d_1(y,x) + d_2(y,x) = (d_1 + d_2)(y,x)$.
Triangle inequality: $(d_1 + d_2)(x,y) = d_1(x,y) + d_2(x,y) \le d_1(x,z) + d_1(z,y) + d_2(x,z) + d_2(z,y) = (d_1 + d_2)(x,z) + (d_1 + d_2)(z,y)$. 
And we are done: $d_1 + d_2$ is a metric on $X$.
Clearly it is not always that simple: there are metrics for which the triangle inequality is not just a one line proof (for example, if $d$ is a metric, $d_1 := \frac{d}{1 + d}$ turns out to be a metric, but you need to work to prove the triangle inequality, or at least you need to work more that we needed to prove that $d_1 + d_2$ is a metric :D )
