# If $d_1, d_2$ are metrics of $X$, is it true that $d_1 +d_2$, $d_1 - d_2$, $d_1\cdot d_2$, $\sqrt d_1$ are metrics on $X$?

If $d_1, d_2$ are metrics of $X$, is it true that $d_1 +d_2$, $d_1 - d_2$, $d_1\cdot d_2$, $\sqrt d_1$ are metrics on $X$?

Here is my attempt:

If we take $d_1 = d_2$ = standard metric on the real line, then $d_1\cdot d_2 = d_1^2$ is not a metric.

$d_1 - d_2$ may not be metric because it may not even be always non-negative.

But I am not sure about others. I need help.

Thanks for giving me time.

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To prove that $d_1+d_2$ is a metric, just check that it has each of the properties of a metric. The only one that takes any work is the triangle inequality, and it’s not hard, using the fact that $d_1$ and $d_2$ both satisfy the triangle inequality. That leaves only $\sqrt{d_1}$, and it’s clear that the only question is whether it satisfies the triangle inequality.
In other words, must it always be true that $$\sqrt{d_1(x,z)}\le\sqrt{d_1(x,y)}+\sqrt{d_1(y,z)}\;?\tag{1}$$
Since both sides of $(1)$ are non-negative, $(1)$ holds iff the inequality that you get by squaring both sides of $(1)$ holds; does it?
Sir, based on your Idea I have done like this: $d_1(x,z)\leq d_1(x,y)+d_1(y,z)$ and $d_2(x,z)\leq d_2(x,y)+d_2(y,z)$ . Adding both will give required result. Thinking about second one. – srijan Jun 4 '12 at 10:11