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Let $X\neq\emptyset$ and let $\rho:X\times X\longrightarrow \mathbb{R}$ with the following properties :

a) for $\forall x,y\in X$, $\rho(x,y)\geq 0$

b) for $\forall x,y\in X$, $\rho(x,y) = 0\iff x=y$.

c) for $\forall x,y,z\in X$, $\rho(x,y)\leq \rho(x,z) + \rho(y,z)$

Show that $\rho$ is a metric on $X$.

I understand that I should show that we can use these three given axioms to reach to the third one.

I substituted firstly $x=z$ and then $y=z$ in the triangle inequality, but really couldn't reach what I want.

If $x=z$, $\rho(x,y)\leq 0+\rho(y,x)$.

If $y=z$, $\rho(x,y)\leq 0+\rho(x,y)$.

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    You only need to prove that $\rho(x,y)=\rho(y,x)$, don't you? – 2017-02-18
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    Yes I know but how – 2017-02-18
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    I was trying to use the given axioms but I couldn't reach to the wanted axiom. – 2017-02-18
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    @user416990: [**Please, look at how I've edited your question to use LaTeX / MathJax properly**](http://math.stackexchange.com/revisions/28c99ab0-90a7-4a6e-a104-03052663a68d/view-source). – 2017-02-18
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    Ok thanks a lot – 2017-02-18

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You had a great idea to let $z=x$, and as you saw, you can conclude from statement (c) that for any $x,y\in X$, $$\rho(x,y)\leq 0+\rho(y,x)$$ But since this is true for any $x$ and $y$, you can use the same $x$ and $y$ just in the other order! $$\rho(y,x)\leq 0+\rho(x,y)$$

Now you have $$\rho(x,y)\leq\rho(y,x)\qquad \rho(y,x)\leq \rho(x,y)$$ and the only way both of those things can be true is if $\rho(x,y)=\rho(y,x)$.

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    Thanks a lot it what I was trying to prove ...thanks – 2017-02-18