Firstly, I regret I can't put the full question in the title part due to character limit. The actual question looks like-
Let, $X$ be a set of all sequences $x=\{x_n\}$ of real numbers and for $x=\{x_n\}$ and $y=\{y_n\}$ of $X$ with $x\ne y$, put $M(x,y)=$min$\{n\in\Bbb{N} : x_n\ne y_n\}$. For any $x, y\in X$, define $$d(x,y) =\begin{cases} 0, & \text{if}\ x=y\\ {1 \over M(x,y)}, & \text{otherwise} \end{cases}$$
It is easy to verify that $d(x,y)\ge 0$, $d(x,y)=d(y,x)\;\forall x,y\in X$ and $d(x,y)=0$ if and only if $x=y$.
So, I have to prove $d(x,z)\le d(x,y)+d(z,y)\ \forall x,z,y\in X\tag{1}\label{eq1}$
Choose $x=\{x_n\},\ y=\{y_n\},\ z=\{z_n\}\in X$
If any two of them are equal then $\eqref{eq1}$ follows directly. Similarly if all of them are equal then it is also obvious.
Assume, $x, y, z$ are the there different sequences.
Let, $M(x,z)=p,\ M(x,y)=q,\ M(z,y)=r$.
My assumption is(not sure) both of $q,r\le p$.
But I cannot write down the proof in a rigorous manner.
Can anybody give me a proper way out for my problem? Thanks for your assistance in advance.
With the notation from the problem, it holds that $p \ge \min\{q,r\}$.
Proof: If $q$ or $r$ are equal to 1, that's obvious. If both are greater, let $s=\min\{q,r\}$. We know that for indices $i=1,2,\ldots,s-1: x_i=y_i$ and $z_i=y_i$, because of the definition of $q$ and $r$ and the definiton of $M(a,b)$. So we get $x_i=z_i$ for $i=1,2,\ldots,s-1$, which means $p=M(x,z) \ge s$.$\blacksquare$
Remark: It can be easily shown that if $q \neq r$ then $p=\min\{q,r\}$.
To prove the triangle inequality needed for $d$ to be a metric (under the assumption $x,y$ and $z$ are mutually different), we simply remark that with $s=\min\{q,r\}$:
$$d(x,z) = \frac1p \le \frac1s < \frac1q + \frac1r = d(x,y) + d(y,z)$$
The sharp inequality follows from the fact that the left hand side ($\frac1s$) is one of the two summands on the right hand side, while the other is positive.