# Show inequality (max, min)

How do I show this inequality

$$d(x,z) \leq \max(d(x,y), d(y,z))$$

when

$$\mu (x,y) = \min\{n\in\mathbb{N} \ | \ x_n \not= y_n \}$$

and

$$d(x,y) = \frac{1}{\mu(x,y)}$$

What I've done so far:

$$d(x,z) = \frac{1}{\mu(x,z)} \Leftrightarrow \mu (x,z) = \frac{1}{d(x,z)} = \min\{n\in\mathbb{N} \ | \ x_n \not= z_n\}$$

Therefore

$$\min\{n\in\mathbb{N} \ | \ x_n \not= z_n\} \leq \min\{\{n\in\mathbb{N} \ | \ x_n \not= y_n\},\{ n\in\mathbb{N} \ | \ y_n \not= z_n \}\}$$

I can't seem to figure out where to go from here.

EDIT: Does any rules apply to the following?

$$\frac{1}{\mu(x,z)} \leq \frac{1}{\min\{ \mu(x,y), \mu(y,z) \}}$$

$x_n, y_n, z_n$ are sequences that are either 0 or 1.

• Right, thanks. Corrected :) – Stranqer95 May 30 '15 at 18:42
• Can you explain what kind of things are $x,y, x_n,y_n$? Real numbers, or vectors and components, or sequences? – Emilio Novati May 30 '15 at 19:10
• Added it to the description. – Stranqer95 May 30 '15 at 19:23

Assume that $\mu(x,y)=n_{1}$ is finite,and $\mu(y,z)=n_{2}$ is finite. Then given, for all $n < n_{1}$ $x_{n}=y_{n}$ and $n<n_{2}$, $y_{n} = z_{n}$, hence for all $n<min\lbrace n_{1},n_{2} \rbrace$ $x_{n} =z_{n}$, so $\mu(x,z) \geq min \lbrace n_{1},n_{2} \rbrace$ sp $d(x,z) \leq max \lbrace \frac{1}{n_{1}},\frac{1}{n_{2}}\rbrace = max \lbrace d(x,y),d(y,z)\rbrace$. Notice that $\frac{1}{min\lbrace n_{1},n_{2} \rbrace}=max\lbrace \frac{1}{n_{1}}, \frac{{1}}{n_{2}}\rbrace$
• Why is $x_n = y_n$ when $n<n_1$? – Stranqer95 May 30 '15 at 19:26
• The definition of $\mu(x,y)$ is the smallest integer at which $x_{n}$ and $y_{n}$ differ! So they all agree at all integers before $n_{1}$. – mich95 May 30 '15 at 19:27
• I don't follow. If $x_n \not= y_n$, the definition of $\mu(x,y)$, then how can $x_n = y_n$? – Stranqer95 May 30 '15 at 19:33
• $\mu(x,y)$ is the smallest integer $n$ for which $x_{n} \not =y_{n}$ eg : $\mu(x,y)=2$ if $x_{n}=n, y_{n}=n^{2}$. Let $n_{1} = \mu(x,y)$. If $n<n_{1}$, since $n_{1}$ is the minimal integer at which $x_{n}$ and $y_{n}$ differ, then for all integers strictly less than it, $x_{n}$ and $y_{n}$ must be equal. – mich95 May 30 '15 at 19:35
• Hm, I'm not sure I know what you mean. I understand that $n_1$ is the minimal integer at which $x_n$ and $y_n$ differ. But I still don't get how $x_n = y_n$. Is it possible to solve it in another way using the method I've written in the description? – Stranqer95 May 30 '15 at 19:54