= Ultrametric inequality
{title2=$d(x,z)\leq\max\{d(x,y),d(y,z)\}$}
The ultrametric inequality, also called the strong <triangle inequality>, bounds a direct distance by the larger of the two distances through an intermediate point:
$$
d(x,z)\leq\max\{d(x,y),d(y,z)\}.
$$
A <metric space> satisfying this rule is an <ultrametric space>. For a <Non-Archimedean absolute value>, the corresponding <inequality> is $|a+b|\leq\max\{|a|,|b|\}$, and the associated <metric> is $d(x,y)=|x-y|$. The ordinary <triangle inequality> follows because the maximum of two nonnegative <real numbers> is at most their <sum>.
For example, the <2-adic absolute value> gives the three points $0,1,8$ distances
$$
d_2(0,1)=1,\qquad d_2(1,8)=|-7|_2=1,\qquad d_2(0,8)=\frac18.
$$
Each distance is bounded by the maximum of the other two. The two largest distances are equal: every triangle in an <ultrametric space> has this property. Otherwise its unique longest side would exceed the maximum of the other two, contradicting the <ultrametric inequality>. The drawn lengths of a triangle on paper need not represent these distances.
The <ultrametric inequality> gives a useful exact rule when $|a|\ne|b|$:
$$
|a+b|=\max\{|a|,|b|\}.
$$
To see why, suppose $|a|>|b|$. The upper bound is $|a+b|\leq|a|$; applying the <ultrametric inequality> to $a=(a+b)-b$ forces the reverse bound. Equal sizes can instead cancel: $|1|_2=|1|_2=1$, but $|1+1|_2=1/2$. This makes a uniquely largest term decisive in a <sum>, a principle used in the <Newton polygon> and in arguments about <roots of a polynomial>.
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