$d: \mathbb{R^2} \times \mathbb{R^2} \rightarrow \mathbb{R}$
where $d(x,y)=\sqrt[3]{|x_1-y_1|+|x_2-y_2|}$ for $x=(x_1, x_2)$ and $y=(y_1, y_2)$
I am trying to determine if $d$ defines a metric on $\mathbb{R^2}$, but am stuck on the triangle inequality condition
The $3$ conditions of a metric are: positivity, symmetry and the triangle inequality
- Positivity. True as: $|x_1-y_1|+|x_2-y_2|$ is positive, and sign is preserved by taking cubed roots.
- Symmetry. True as: $d(y,x)=\sqrt[3]{|y_1-x_1|+|y_2-x_2|}=\sqrt[3]{|(-1)(-y_1+x_1)|+|(-1)(-y_2+x_2)|}=\sqrt[3]{|x_1-y_1|+|x_2-y_2|}=d(x,y)$
- Triangle inequality. Need to show $d(x, y)+d(y,z) \leq d(x, z) $
$d(x,y)+d(y,z)=\sqrt[3]{|x_1-y_1|+|x_2-y_2|}+\sqrt[3]{|y_1-z_1|+|y_2-z_2|}$
$d(x,z)=\sqrt[3]{|x_1-z_1|+|z_2-y_2|}$
I am having trouble combining them to prove the inequality.
I have also tried finding possible counterexamples but to no avail
Would really appreciate your help, thanks