# Does $d(x,y)=\sqrt[3]{|x_1-y_1|+|x_2-y_2|}$ define a metric?

$d: \mathbb{R^2} \times \mathbb{R^2} \rightarrow \mathbb{R}$

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

1. Positivity. True as: $|x_1-y_1|+|x_2-y_2|$ is positive, and sign is preserved by taking cubed roots.
2. Symmetry. True as: $d(y,x)=\sqrt[3]{|y_1-x_1|+|y_2-x_2|}=\sqrt[3]{|(-1)(-y_1+x_1)|+|(-1)(-y_2+x_2)|}=\sqrt[3]{|x_1-y_1|+|x_2-y_2|}=d(x,y)$
3. 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

If $A^3 \ge B^3$ then $A \ge B$.

$d(x,y)+d(y,z)=\sqrt[3]{|x_1-y_1|+|x_2-y_2|}+\sqrt[3]{|y_1-z_1|+|y_2-z_2|}$

$d(x,z)=\sqrt[3]{|x_1-z_1|+|z_2-y_2|}$

$(d(x,y)+d(y,z))^3 = |x_1-y_1|+|x_2-y_2|+|y_1-z_1|+|y_2-z_2|+3(d(x,y)+d(y,z))\ge$ $\ge|x_1-z_1|+|z_2-y_2|=d(x,z)^3$.

Then $d(x,y)+d(y,z) \ge d(x,z)$

So $d(x,y)+d(y,z)\ge 0$, then $$|x_1-y_1|+|x_2-y_2|+|y_1-z_1|+|y_2-z_2|+3(d(x,y)+d(y,z))\ge$$ $$\ge |x_1-y_1|+|x_2-y_2|+|y_1-z_1|+|y_2-z_2|$$
So $|a|+|b|\ge |a+b|$, then $$|x_1-y_1|+|x_2-y_2|+|y_1-z_1|+|y_2-z_2|\ge$$ $$\ge |x_1-y_1+x_2-y_2|+|y_1-z_1+y_2-z_2|=$$ $$=|x_1-z_1|+|x_2-z_2|$$
• Ah yes, taking cubes seems the good way to go. I have looked through your solution and can't see how you know $|x_1-y_1|+|x_2-y_2|+|y_1-z_1|+|y_2-z_2|+3(d(x,y)+d(y,z)) \geq |x_1-z_1|+|z_2-y_2|$. Could you give me an idea as this is not trivial to me. Many thanks – thinker Apr 1 '16 at 17:06