why does we need to satisfy $d(x,y)\le d(x,z)+d(z,y)$ in order to show metric space? In metric space some axioms must be satisfied . I wonder why we need to satisfy  $d(x,y)\le d(x,z)+d(z,y)$  in order to be metric space. If this axiom is not satisfied, does any problems occur?
    Thank you for your helping
 A: A metric space is meant to formalize some notion of distance. That axiom states that we never get a situation like:

Well, it's a long ways from point $A$ to point $C$ - but the total journey is way shorter if we stop by $B$ first!

Which is clearly outside of our intuition for how distances should work - adding more way points shouldn't decrease the distance - if it does, we should have chosen that path to measure our distance by in the first place.
This causes serious theoretical problems. In particular, let $B(x,r)$ be a ball of radius $x$ around $r$:
$$B(x,r)=\{y:d(x,y)< r\}.$$
 We can define an open set $O$ to be one such that, for each $x\in O$, there is a $r$ such that $B(x,r)$ is contained in $O$. Ordinarily, you can prove that a ball is open since, for any $y\in B(x,r)$ we have that the ball $B(y,r-d(x,y))$ is contained in $B(x,r)$ because for any $z\in B(y,r-d(x,y))$ we have:
$$d(y,z)<r-d(x,y)$$
$$d(x,y)+d(y,z)<r$$
And then we use the triangle inequality to prove
$$d(x,z)\leq d(x,y)+d(y,z)<r$$
implying $z$ is in $B(x,r)$.
That axiom gives us the machinery to show that open balls are actually open - which is kind of important. In fact, if we take the usual metric on $\mathbb R$, but consider the distance from $0$ to $n$ for natural $n$ to be $\frac{1}n$, which violates the triangle inequality, then any ball of finite radius not centered on $0$, but containing $0$ is not open, because any ball around $0$ contains arbitrarily large integers, but any ball of finite radius not centered on $0$ does not. Once we've lost the statement that open balls are open, any topology we wanted to do is going to be way harder.
We can, however, salvage some non-metrics $d$ violating only the triangle inequality by defining
$$d'(x,y)=\inf\{d(x,s_1)+d(s_1,s_2)+\ldots+d(s_{n-1},s_n)+d(s_n,y):s_i\in M\}$$
that is, defining a metric to be the "shortest path" under the old non-metric. This satisfies the triangle inequality, but it might have $d(x,y)=0$ for $x\neq y$.
A: The problem is that we often use it in proofs. To understand better which is the issue, consider the following simple metric:
On $\mathbb R$ define ($d(x,y)= |x-y|$ if $x-y=0$ or $x-y$ is irrational) and $d(x,y)=1$ if $x \neq y$ and $x-y \in \mathbb Q$.
This $d$ satisfies all the requirements of metric, excepting the triangle inequality. 
Here are just couple of the minor issues which appear in this metric.
Issue 1: $x_{n}=\frac{\sqrt{2}}{n}$ and $y_n=\frac{1}{n}-\frac{\sqrt{2}}{n}$ are convergent, but $x_n+y_n$ is not.
The metric can actually be changed so that you can have $x_n$ convergent but $x_n+x_n$ divergent. 
Issue 2: The sequence $x_{2n}=\frac{\sqrt{2}}{n} \,;\, x_{2n+1}=\frac{1}{n}+\frac{\sqrt{2}}{n}$ converges to $0$ but it is not Cauchy. 
A: Usually  a metric is defined on a vector space $S$ as a function $d: S\times S \rightarrow \mathbb{R}$ such that $\forall x,y,z,\in S$:
$$
1) \qquad d(x,y)\ge 0 \land d(x.,y)= \iff x=y
$$
$$
2) \qquad d(x,y)=d(y,x)
$$
$$
3) \qquad d(x,z)\le d(x,y)+d(y,z)
$$
The condition $1)$ means that $d$ is a positive defined function and the $3)$ is the triangle inequality of OP. 
This condition incorporates our intuitive knowledge that the ''distance'' between two points is the minimum length of the path that we have to done from one point to the other. And in our everyday experience this '' minimum path'' is a straight line segment between the two point.
More! As noted in the other answers, if condition $3)$ is not satisfied, we have some trouble in defining a topology on the space, i.e. to give a good definition of the notion of ''closeness'' of points.
There are many ways of relaxing conditions $1),2),3)$ ( see here)  that, as far as i know, are useful mainly in functional analysis. But this is a too intricate world.
I think that it's more interesting another question: what happens if we live in a space that is not ''flat''? Our Earth is a sphere and we know that the minimum path between two point is not a segment of straight line, but an arc of a maximum circle. This bring us to a more general definition of distance, as a sum of many little segments between very close points (an integral at the limit) , and this distance is the minimum path between the points because we suppose that locally, i.e. between two very close points, the distance is well approximated by a segment of straight line, and the triangle inequality is valid. This simple consideration is incorporated in the magnificent theory of Riemannian manifolds where the metric is defined by means of a positive defined quadratic form ( see here).
But we can relax also this condition and, if the metric is not a positive defined form, but only a non-degenerate one, we meet pseudo-riemannian manifolds.  Specially important are the Lorentz manifolds in which there are points such that the ''distance'' between which is not the minimum length of the possible paths, but the maximum length. 
 E.g.: for a two dimensional Lorentz manifold the distance from two near points is not $ds^2=dx^2+dy^2$, but $ds^2=dx^2-dy^2$.
This means that there are distinct points from which the distance is null, and  the triangle inequality is inverted: $ d(x.y)\ge d(x,y)+d(y,z)$.
This is the geometry of the Theory of Relativity, that seems counter-intuitive and leads to many well-known ''paradoxes''. But it seems that this is the true geometry of space-time in which we live.
