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For $|\cdot|_a:\mathbb{Q}\rightarrow\mathbb{R}$, with the property $$|x+y|_a\leq\max\{|x|_a,|y|_a\},$$ define $d:\mathbb{Q}\times\mathbb{Q}\rightarrow\mathbb{R}$ with $d(x,y)=|x+y|_a$.

How do we show that $d(x,z)\leq d(x,y)+d(y,z)$?

We know that $|x+z|_a\leq\max\{|x|_a,|z|_a\}$, but how does this give the claim?

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    If $\lvert x\rvert_a < 0$ is possible, it need not be true. – 2017-02-04
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    Whats the definition of Norm here ? – 2017-02-04
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    You probably meant $d(x,y) = |x-y|_a$, not the sum. How else do we get $d(x,x) = 0$? – 2017-02-05

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If the distance $d$ is defined as $d(x,y) = \| x - y\|$, which makes more sense to me, then you can do the following: $$\|x-z\| = \|x-y+y-z\| \leq \max\{\|x-y\|,\|y-z\|\} \leq \|x-y\|+\|y-z\|$$

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    Why is the max $\geq||x-y+y-z||$? – 2017-02-04
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    By the property of the distance – 2017-02-04
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I assume you meant $d(x,y) = |x-y|_a$ otherwise we don't have a metric.

If $x,y,z \in \mathbb{Q}$, $x-z =(x-y) + (y- z)$, the $y's$ cancel out, so

$$d(x,z) = |(x-z)|_a = \left|(x-y) + (y- z)\right|_a \le \max(|x-y|_a, |y-z|_a) = \max(d(x,y), d(y,z)) \le d(x,y) + d(x,z)$$

by the property of the norm $|\cdot|_a$ and the fact that the maximum of two numbers is $\le$ than their sum of both are $\ge 0$ (it's going to be a strict $<$ unless one of the numbers is $0$).