I am looking to prove whether or not the following statement is true:
Let $M$ be a set and $D_1$ and $D_2$ metrics on M such that they form a metric space. Then $(M, \min(D_1 , D_2))$ forms a metric space.
Proof: By the definiton of a metric space, the above must satisfy the triangle inequality, or, for some $a,b,c \in M$:
$$ \min [ D_1 (a,c) , D_2 (a,c) ] \le \min [ D_1 (a,b) , D_2 (a,b) ] + \min [ D_1 (b,c) , D_2 (b,c) ] $$
Suppose $\min [ D_1 (a,c) , D_2 (a,c) ] = D_1(a,c)$, and suppose that at least one of the $\min$ functions on the right-hand side is equal to $D_2$, then the triangle-inequality may or may not hold.
From here, would my proof be considered complete?
Or, would I be better providing further clarification, such as a counter-example?
Assume some values for each of the $D$'s that satisfy the triangle inequality for each individually and insert them into the triangle inequality.
$$ \min [ 2 , 5 ] \le \min [ 2 , 1 ] + \min [ 0 , 4 ] $$
This inequality is false, and thus their exists some values such that $(M, \min(D_1 , D_2))$ does not form a metric space.