# Does $d(x+u, y + v) \le d(x, y) + d(u,v)$ holds for every metric?

The title said it, I want to prove that $$d(x+u, y + v) \le d(x, y) + d(u,v)$$ for every metric $d$. If the metric is induced by a norm, i.e. $d(x,y) := ||x-y||$, then this is easy. \begin{align*} d(x+u, y+v) & = ||x+u - (y+v)|| \\ & = ||x-y + u - v|| \\ & \le ||x-y||+||u-v|| \\ & = d(x,y) + d(u,v) \end{align*} But in the general case I have no idea how to get rid of the sums...

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What is the underlying space here? –  user27126 Jan 22 '13 at 19:51
$x+u$ isn’t meaningful in metric spaces in general; are you working specifically in $\Bbb R^n$? –  Brian M. Scott Jan 22 '13 at 19:51

It isn’t even true in $\Bbb R$. The function

$$d:\Bbb R^2\to\Bbb R:\langle x,y\rangle\mapsto\left|\tan^{-1}x-\tan^{-1}y\right|$$

is a metric on $\Bbb R$. In this metric we have $d(-1,1)=\frac{\pi}2$, but $$\lim_{x\to\infty}d(x-1,x+1)=\lim_{x\to\infty}\left|\tan^{-1}(x-1)-\tan^{-1}(x+1)\right|=0\;,$$

even though $d(x,x)=0$ for all $x$.

More generally, suppose that it were true. Then you’d have $d(x+u,y+u)\le d(x,y)$ for all $x,y,u$, and hence $d(x,y)\le d\big((x+u)+(-u),(y+u)+(-u)\big)\le d(x+u,y+u)$ for all $x,y$, and $d$ would be translation-invariant. Clearly not all metrics on $\Bbb R^n$ are translation-invariant.

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in what manner does your limit for $x\to \infty$ said that $d(x-1,x+1) \le d(x,x) + d(-1,1)$ could not hold? –  Stefan Jan 22 '13 at 20:07
@Stefan: It doesn’t: it shows that for sufficiently large $x$ the inequality $$d\big((x-1)+(-x),(x+1)+(-x)\big)\le d(x-1,x+1)+d(-x,-x)$$ cannot hold. –  Brian M. Scott Jan 22 '13 at 20:13

If I'm not mistaken, you can also take the pullback of the standard metric on $ℝ$ by $x ↦ x^3$, i.e. the metric $ℝ × ℝ → ℝ,\; (x,y) ↦ |x^3 - y^3|$, and choose $x=u=1$ and $y=v=-1$. Then the left hand side of the title inequality is $16$ and the right hand side is $2 + 2 = 4$.

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