Is d(x,y) = |x^3 - y^3| a Valid Metric in X?

[mynameisfunk]

Prove or disprove d is a metrix in $$X$$:
$$d(x,y)=|x^3-y^3|$$

OK, 3 conditions to meet:

(i) $$d(x,y)>0$$
(ii) $$d(x,y)=0$$ iff $$x=y$$
(iii) $$d(x,y) \leq d(x,r) + d(r,y)$$ for any $$r \epsilon X$$

the first 2 are obvious and I have solved this by proving all of the cases:

$$r \leq x \leq y$$, $$x \leq r \leq y$$, etc.

My problem is that I know there is a better proof that is much shorter. My professor did it in class but I still had gotten the problem correct so didnt write it down. Any suggestions on a simpler way to do this?

 

[joshyxc1979]

iii is applying the triangle inequality for real numbers, which is a metric. here is a proof of that: http://math.ucsd.edu/~wgarner/math4c/derivations/other/triangleinequal.htm

 

[HallsofIvy]

I think joshyxc1979 has misunderstood the question. The link he provides gives a proof that the usual metric, d(x,y)= |x- y|, satisfies the triangle inequality. That says nothing about whether this new metric satisfies it.

mynameisfunk, try using the fact that [itex]|x^2- y^3|= |x- y||x^2+ xy+ y^2|[itex].

 

[mynameisfunk]

ok.
$$|x^3-y^3| \leq |x^3-r^3| + |r^3-y^3|$$ gives
$$|x-y||x^2+xy+xy^2| \leq |x-r||x^2+xr+r^2| + |r-y||r^2+ry+y^2|$$
since we know the usual metric |x-y| holds, we need to show that,
$$|x^2+xy+y^2| \leq |x^2+xr+r^2| + |r^2+ry+y^2|$$
Now I am stuck, I am not sure If I am allowed to turn this into
$$|xy| \leq |xr|+|ry|$$?
If I did, this wouldn't eve hold true. Also, since it is not true, I cannot add $$|xy| |xr| |ry|$$ to the terms to get squares...

 

[deluks917]

For anyone still looking for the solution:

|a + b| <= |a| + |b|

d(x,y) = |x^3 + y^3| = |(x^3 - r^3) + (r^3 - y^3)| <= |x^3 - r^3| + |r^3 - y^3| =
d(x,r) + d(r,y)