Jaccard dissimilarity and the triangle inequality Suppose that $\delta(A, B) = \dfrac{|A \Delta B|}{|A \cup B|}$, where $\Delta$ represents symmetric difference. Then how does one prove the triangle inequality, viz that $\delta(A, B) + \delta(B, C) \ge \delta(A, C)$?
 A: If I understand you correctly, the objects you are considering are finite sets and your dissimilarity function is defined as $d(A, B) = \frac{|A \ominus B|}{|A \cup B|}$, where $\ominus$ denotes the symmetric difference.
Consider the Venn diagram for three sets $A$, $B$, and $C$.

Then the triangle inequality boils down to
$$
\frac{a+b+e+f}{a+b+d+e+f+g} + \frac{b+c+d+f}{b+c+d+e+f+g} \geq \frac{a+c+d+e}{a+c+d+e+f+g}
$$
Plugging this into Wolfram Alpha, we see that it "simplifies" to:
$$
\frac{a^2 b + a b^2 + a^2 c + 2 a b c + b^2 c + a c^2 + b c^2 + a^2 d + 2 a b d + b^2 d + 2 a c d + 2 b c d + a d^2 + b d^2 + 2 a b e + b^2 e + 2 a c e + 2 b c e + c^2 e + a d e + 2 b d e + c d e + b e^2 + c e^2 + a^2 f + 4 a b f + 2 b^2 f + 4 a c f + 4 b c f + c^2 f + 4 a d f + 5 b d f + 3 c d f + 2 d^2 f + 3 a e f + 5 b e f + 4 c e f + 4 d e f + 2 e^2 f + 3 a f^2 + 4 b f^2 + 3 c f^2 + 4 d f^2 + 4 e f^2 + 2 f^3 + 2 a b g + 2 b^2 g + 2 a c g + 2 b c g + a d g + 3 b d g + 3 b e g + c e g + 3 a f g + 6 b f g + 3 c f g + 4 d f g + 4 e f g + 4 f^2 g + 2 b g^2 + 2 f g^2}{(a + b + d + e + f + g) (a + c + d + e + f + g) (b + c + d + e + f + g)} \geq 0
$$
Since all the coefficients in the above expression are positive, and $a, b, c, d, e, f, g \geq 0$, we conclude that the inequality always holds.
