Showing that product of two metrics is not always a metric 
Let $d_1$ and $d_2$ be two metrics on space $X$. Is $d_1\cdot d_2$ metric on space $X$?

I know that is satisfies first three properties of metric. How do I show that triangle inequality holds or does not hold (an example) for this?
 A: Consider the real line $\mathbb R$ with metrics $d_1(x,y)=|x-y|$ and $d_2(x,y)=\min\{1,d_1(x,y)\}$. The triangle inequality should hold for $d_1\cdot d_2$ for all points $x,y,z\in\mathbb R$ for it to be a metric.
Consider the points $x=1,y=2,z=2.5$ we have that -
$d_1(1,2.5)\cdot d_2(1,2.5)=|1-2.5|\cdot\min\{1,|1-2.5|\}=1.5\cdot 1=1.5$
$d_1(1,2)\cdot d_2(1,2)+d_1(2,2.5)\cdot d_2(2,2.5)=1\cdot1+0.5\cdot0.5=1.25<1.5$
Thus triangle inequality doesn't hold.
A: The product of two metrics may or may not be a metric. I'll give two examples for these.
When the product is a metric
Consider $d_1, d_2$ be the discrete metric on $X$, then $d_1 \cdot d_2$ is again a discrete metric and so the product is a metric.
When the product is not a metric
Consider $X= \mathbb{R}$ with $d_1 = d_2$ usual metric, $i.e.$ $d_1(x,y)= |x-y|$, and $d=d_1\cdot d_2$.
Then the triangle inequality is not satisfied as

*

*$d(0,1)= 1^2$,

*$d(0,1/2)= 1/2^2$,

*$d(1/2,1)=1/2^2$.

and so $d(0,1) \nleq d(0,1/2)+ d(1/2,1)$, hence $d$ is not a metric.
A: I went over a couple of exercises like this and it's usually enough to just draw a triangle in order to look for counter-examples. Instead of defining metrics with a formula, define them by just listing their value on the appropriate edge:

