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I'm looking for the proof of that: d(x,y) = d (y,x). I know that I have to use the "non-negativity" and "triangle inequality" but I don't know how to combine them to get the result.

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In a metric space, this is part of the definition, so not something that can be proved. – Tobias Kildetoft Oct 8 '13 at 18:09
However, you may be asked to prove that some distance function $d$ is a metric, in which case you would need to verify the symmetry property. By the way, symmetry does not follow from nonnegativity plus triangle inequality; why do you think it does? – user43208 Oct 8 '13 at 18:12
I want to proove that the symmetry follows from nonnegativity and triangle inequality or not. – Alex Oct 8 '13 at 18:14

1 Answer 1

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Take $d(x,y)=x^2+2y^2$. We obviously have nonnegativity and $d(x,y)+d(y,z)=x^2+2y^2+y^2+2z^2\geq x^2+2z^2=d(x,z)$ so the triangle inequality holds. But, for instance, $d(1,0)=1$ and $d(0,1)=2$, so symmetry does not hold. So symmetry does not follow from nonnegativity and the triangle inequality.

As mentioned in the comments, symmetry is a part of the definition of a distance function. A distance function is a real-valued function which satisfies nonnegativity, symmetry, and the triangle inequality, as well as $d(x,y)=0\iff x=y$. You can come up with functions which satisfy any three of these and do not satisfy the fourth—that's why they're all required in the definition.

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