linguistic:paraphrasing
Reverse triangle inequality

The reverse triangle inequality is a logically equivalent alternative formulation of the triangle inequality that gives lower bounds instead of upper bounds. For plane geometry, the statement is:

In the case of a normed vector space, the statement is:

or for metric spaces, | d (A, C) − d (B, C) | ≤ d (A, B) {\displaystyle |d(A,C)-d(B,C)|\leq d(A,B)}.This implies that the norm ‖ ⋅ ‖ {\displaystyle \|\cdot \|} as well as the distance-from- z {\displaystyle z} function d (z, ⋅) {\displaystyle d(z,\cdot)} are Lipschitz continuous with Lipschitz constant 1, and therefore are in particular uniformly continuous.

The proof of the reverse triangle inequality from the usual one uses ‖ v − u ‖ = ‖ − 1 (u − v) ‖ = | − 1 | ⋅ ‖ u − v ‖ = ‖ u − v ‖ {\displaystyle \|v-u\|=\|{-}1(u-v)\|=|{-}1|\cdot \|u-v\|=\|u-v\|} to find:

Combining these two statements gives:

In the converse, the proof of the triangle inequality from the reverse triangle inequality works in two cases:

If ‖ u + v ‖ − ‖ u ‖ ≥ 0, {\displaystyle \|u+v\|-\|u\|\geq 0,} then by the reverse triangle inequality, ‖ u + v ‖ − ‖ u ‖ = | ‖ u + v ‖ − ‖ u ‖ | ≤ ‖ (u + v) − u ‖ = ‖ v ‖ ⇒ ‖ u + v ‖ ≤ ‖ u ‖ + ‖ v ‖ {\displaystyle \|u+v\|-\|u\|={\big |}\|u+v\|-\|u\|{\big |}\leq \|(u+v)-u\|=\|v\|\Rightarrow \|u+v\|\leq \|u\|+\|v\|},

and if ‖ u + v ‖ − ‖ u ‖ < 0, {\displaystyle \|u+v\|-\|u\|<0,} then trivially ‖ u ‖ + ‖ v ‖ ≥ ‖ u ‖ > ‖ u + v ‖ {\displaystyle \|u\|+\|v\|\geq \|u\|>\|u+v\|} by the nonnegativity of the norm.

Thus, in both cases, we find that ‖ u ‖ + ‖ v ‖ ≥ ‖ u + v ‖ {\displaystyle \|u\|+\|v\|\geq \|u+v\|}.

For metric spaces, the proof of the reverse triangle inequality is found similarly by:

d (A, B) + d (B, C) ≥ d (A, C) ⇒ d (A, B) ≥ d (A, C) − d (B, C) {\displaystyle d(A,B)+d(B,C)\geq d(A,C)\Rightarrow d(A,B)\geq d(A,C)-d(B,C)}

d (C, A) + d (A, B) ≥ d (C, B) ⇒ d (A, B) ≥ d (B, C) − d (A, C) {\displaystyle d(C,A)+d(A,B)\geq d(C,B)\Rightarrow d(A,B)\geq d(B,C)-d(A,C)}

Putting these equations together we find:

d (A, B) ≥ | d (A, C) − d (B, C) | {\displaystyle d(A,B)\geq |d(A,C)-d(B,C)|}

And in the converse, beginning from the reverse triangle inequality, we can again use two cases:

If d (A, C) − d (B, C) ≥ 0 {\displaystyle d(A,C)-d(B,C)\geq 0}, then d (A, B) ≥ | d (A, C) − d (B, C) | = d (A, C) − d (B, C) ⇒ d (A, B) + d (B, C) ≥ d (A, C) {\displaystyle d(A,B)\geq |d(A,C)-d(B,C)|=d(A,C)-d(B,C)\Rightarrow d(A,B)+d(B,C)\geq d(A,C)},

and if d (A, C) − d (B, C) < 0, {\displaystyle d(A,C)-d(B,C)<0,} then d (A, B) + d (B, C) ≥ d (B, C) > d (A, C) {\displaystyle d(A,B)+d(B,C)\geq d(B,C)>d(A,C)} again by the nonnegativity of the metric.

Thus, in both cases, we find that d (A, B) + d (B, C) ≥ d (A, C) {\displaystyle d(A,B)+d(B,C)\geq d(A,C)}.