Triangle inequality

When {{mvar|r}} the common ratio is chosen such that {{math|r {{=}} {{sqrt|φ}}}} it generates a right triangle that is always similar to the Kepler triangle.

:$\backslash begin\{array\}\{rcccl\}$

0 &<& a &<& ar+ar^2+ar^3 \\

0 &<& ar &<& a+ar^2+ar^3 \\

0 &<& ar^2 &<& a+ar+ar^3 \\

0 &<& ar^3 &<& a+ar+ar^2.

\end{array}

These inequalities for {{math|a > 0}} reduce to the following

:$r^3+r^2+r-1>0$

:{{cite journal|title=input: solve 02+ar³, 032 |last=Wolfram{{!}}Alpha|journal=Wolfram Research|url=http://www.wolframalpha.com/input/?i=solve%20{0%3Ca%3Ca*r%2Ba*r^2%2Ba*r^3%2C%200%3Ca*r^3%3Ca%2Ba*r%2Ba*r^2}&t=ff3tb01|access-date=2012-07-29}}

The left-hand side polynomials of these two inequalities have roots that are the tribonacci constant and its reciprocal. Consequently, {{mvar|r}} is limited to the range {{math|1/t < r < t}} where {{mvar|t}} is the tribonacci constant.

This generalization can be used to prove that the shortest curve between two points in Euclidean geometry is a straight line.

No polygonal path between two points is shorter than the line between them. This implies that no curve can have an arc length less than the distance between its endpoints. By definition, the arc length of a curve is the least upper bound of the lengths of all polygonal approximations of the curve. The result for polygonal paths shows that the straight line between the endpoints is the shortest of all the polygonal approximations. Because the arc length of the curve is greater than or equal to the length of every polygonal approximation, the curve itself cannot be shorter than the straight line path.{{cite book|title=Numbers and Geometry|author=John Stillwell|author-link=John Stillwell|year=1997|publisher=Springer|isbn=978-0-387-98289-2|url=https://books.google.com/books?id=4elkHwVS0eUC&pg=PA95}} p. 95.

In either case, if the side lengths are {{mvar|a}}, {{mvar|b}}, {{mvar|c}} we can attempt to place a triangle in the Euclidean plane as shown in the diagram. We need to prove that there exists a real number {{mvar|h}} consistent with the values {{mvar|a}}, {{mvar|b}}, and {{mvar|c}}, in which case this triangle exists.

Image:Triangle with notations 3.svg

By the Pythagorean theorem we have {{math|b{{sup|2}} {{=}} h{{sup|2}} + d{{sup|2}}}} and {{math|a{{sup|2}} {{=}} h{{sup|2}} + (c − d){{sup|2}}}} according to the figure at the right. Subtracting these yields {{math|a{{sup|2}} − b{{sup|2}} {{=}} c{{sup|2}} − 2cd}}. This equation allows us to express {{mvar|d}} in terms of the sides of the triangle:

:$d=\backslash frac\{-a^2+b^2+c^2\}\{2c\}.$

For the height of the triangle we have that {{math|h{{sup|2}} {{=}} b{{sup|2}} − d{{sup|2}}}}. By replacing {{mvar|d}} with the formula given above, we have

:$h^2\; =\; b^2-\backslash left(\backslash frac\{-a^2+b^2+c^2\}\{2c\}\backslash right)^2.$

For a real number {{mvar|h}} to satisfy this, {{math|h{{sup|2}}}} must be non-negative:

:$\backslash begin\{align\}$

0 &\le b^2-\left(\frac{-a^2+b^2+c^2}{2c}\right)^2 \\[4pt]

0 &\le \left(b- \frac{-a^2+b^2+c^2}{2c}\right) \left(b + \frac{-a^2+b^2+c^2}{2c}\right) \\[4pt]

0 &\le \left(a^2-(b-c)^2)((b+c)^2-a^2 \right) \\[6pt]

0 &\le (a+b-c)(a-b+c)(b+c+a)(b+c-a) \\[6pt]

0 &\le (a+b-c)(a+c-b)(b+c-a)

\end{align}

which holds if the triangle inequality is satisfied for all sides. Therefore, there does exist a real number $h$ consistent with the sides $a,\; b,\; c$, and the triangle exists. If each triangle inequality holds strictly, $h\; >\; 0$ and the triangle is non-degenerate (has positive area); but if one of the inequalities holds with equality, so $h\; =\; 0$, the triangle is degenerate.

{{see also|Distance geometry#Cayley–Menger determinants}}

The area of a triangular face of a tetrahedron is less than or equal to the sum of the areas of the other three triangular faces. More generally, in Euclidean space the hypervolume of an {{math|(n − 1)}}-facet of an {{mvar|n}}-simplex is less than or equal to the sum of the hypervolumes of the other {{mvar|n}} facets.

Much as the triangle inequality generalizes to a polygon inequality, the inequality for a simplex of any dimension generalizes to a polytope of any dimension: the hypervolume of any facet of a polytope is less than or equal to the sum of the hypervolumes of the remaining facets.

In some cases the tetrahedral inequality is stronger than several applications of the triangle inequality. For example, the triangle inequality appears to allow the possibility of four points {{mvar|A}}, {{mvar|B}}, {{mvar|C}}, and {{mvar|Z}} in Euclidean space such that distances

:{{math|1=AB = BC = CA = 26}}

and

:{{math|1=AZ = BZ = CZ = 14}}.

However, points with such distances cannot exist: the area of the {{math|26–26–26}} equilateral triangle {{math|ABC}} is $169\backslash sqrt\; 3$, which is larger than three times $39\backslash sqrt\; 3$, the area of a {{math|26–14–14}} isosceles triangle (all by Heron's formula), and so the arrangement is forbidden by the tetrahedral inequality.

In a normed vector space {{mvar|V}}, one of the defining properties of the norm is the triangle inequality:

:$\backslash |\backslash mathbf\; u\; +\; \backslash mathbf\; v\backslash |\; \backslash leq\; \backslash |\backslash mathbf\; u\backslash |\; +\; \backslash |\backslash mathbf\; v\backslash |\; \backslash quad\; \backslash forall\; \backslash ,\; \backslash mathbf\; u,\; \backslash mathbf\; v\; \backslash in\; V$

That is, the norm of the sum of two vectors is at most as large as the sum of the norms of the two vectors. This is also referred to as subadditivity. For any proposed function to behave as a norm, it must satisfy this requirement.

{{cite book |title=Numerical analysis |author=Rainer Kress |chapter=§3.1: Normed spaces |chapter-url=https://books.google.com/books?id=e7ZmHRIxum0C&pg=PA26 |page=26 |isbn=0-387-98408-9 |year=1988 |publisher=Springer}}

If the normed space is Euclidean, or, more generally, strictly convex, then $\backslash |\backslash mathbf\; u+\backslash mathbf\; v\backslash |=\backslash |\backslash mathbf\; u\backslash |+\backslash |\backslash mathbf\; v\backslash |$ if and only if the triangle formed by {{math|u}}, {{math|v}}, and {{math|u + v}}, is degenerate, that is, {{math|u}} and {{math|v}} are on the same ray, i.e., {{math|u {{=}} 0}} or {{math|v {{=}} 0}}, or {{math|u {{=}} α v}} for some {{math|α > 0}}. This property characterizes strictly convex normed spaces such as the {{math|ℓ_p}} spaces with {{math|1 < p < ∞}}. However, there are normed spaces in which this is not true. For instance, consider the plane with the {{math|ℓ₁}} norm (the Manhattan distance) and denote {{math|u {{=}} (1, 0)}} and {{math|v {{=}} (0, 1)}}. Then the triangle formed by {{math|u}}, {{math|v}}, and {{math|u + v}}, is non-degenerate but

:$\backslash |\backslash mathbf\; u+\backslash mathbf\; v\backslash |=\backslash |(1,1)\backslash |=|1|+|1|=2=\backslash |\backslash mathbf\; u\backslash |+\backslash |\backslash mathbf\; v\backslash |.$

$$|u\; +\; v|\; \backslash leq\; |u|+|v|.$$

The triangle inequality is useful in mathematical analysis for determining the best upper estimate on the size of the sum of two numbers, in terms of the sizes of the individual numbers. There is also a lower estimate, which can be found using the reverse triangle inequality which states that for any real numbers {{mvar|u}} and {{mvar|v}}, $|u-v|\; \backslash geq\; \backslash bigl||u|-|v|\backslash bigr|.$

The taxicab norm or 1-norm is one generalization absolute value to higher dimensions. To find the norm of a vector $v\; =\; (v\_1,\; v\_2,\; \backslash ldots\; v\_n\; ),$ just add the absolute value of each component separately, $$\backslash |v\backslash |\_1\; =\; |v\_1|\; +\; |v\_2|\; +\; \backslash dotsb\; +\; |v\_n|.$$

The Euclidean norm or 2-norm defines the length of translation vectors in an {{mvar|n}}-dimensional Euclidean space in terms of a Cartesian coordinate system. For a vector $v\; =\; (v\_1,\; v\_2,\; \backslash ldots\; v\_n\; ),$ its length is defined using the {{mvar|n}}-dimensional Pythagorean theorem: $$\backslash |v\backslash |\_2\; =\; \backslash sqrt$$

The Cauchy–Schwarz inequality turns into an equality if and only if {{mvar|u}} and {{mvar|v}} are linearly dependent. The inequality $\backslash langle\; u,\; v\; \backslash rangle\; +\; \backslash langle\; v,\; u\; \backslash rangle\; \backslash le\; 2\backslash left|\backslash left\backslash langle\; u,\; v\; \backslash right\backslash rangle\backslash right|$ turns into an equality for linearly dependent $u$ and $v$ if and only if one of the vectors {{mvar|u}} or {{mvar|v}} is a nonnegative scalar of the other. Taking the square root of the final result gives the triangle inequality.

The p-norm is a generalization of taxicab and Euclidean norms, using an arbitrary positive integer exponent, $$\backslash |v\backslash |\_p\; =\; \backslash bigl(|v\_1|^p\; +\; |v\_2|^p\; +\; \backslash dotsb\; +\; |v\_n|^p\backslash bigr)^\{1/p\},$$ where the {{math|v_i}} are the components of vector {{mvar|v}}.

Except for the case {{math|1=p = 2}}, the {{mvar|p}}-norm is not an inner product norm, because it does not satisfy the parallelogram law. The triangle inequality for general values of {{mvar|p}} is called Minkowski's inequality.{{cite book |title=Beginning functional analysis |author= Karen Saxe|author-link= Karen Saxe |url=https://books.google.com/books?id=0LeWJ74j8GQC&pg=PA61 |page=61 |isbn=0-387-95224-1 |publisher=Springer |year=2002}} It takes the form:$$\backslash |u+v\backslash |\_p\; \backslash le\; \backslash |u\backslash |\_p\; +\; \backslash |v\backslash |\_p\; \backslash \; .$$

:$d(A,\backslash \; C)\; \backslash le\; d(A,\backslash \; B)\; +\; d(B,\backslash \; C)\; \backslash \; ,$

for all points {{mvar|A}}, {{mvar|B}}, and {{mvar|C}} in {{mvar|M}}. That is, the distance from {{mvar|A}} to {{mvar|C}} is at most as large as the sum of the distance from {{mvar|A}} to {{mvar|B}} and the distance from {{mvar|B}} to {{mvar|C}}.

The triangle inequality is responsible for most of the interesting structure on a metric space, namely, convergence. This is because the remaining requirements for a metric are rather simplistic in comparison. For example, the fact that any convergent sequence in a metric space is a Cauchy sequence is a direct consequence of the triangle inequality, because if we choose any {{math|x_n}} and {{math|x_m}} such that {{math|d(x_n, x) < ε/2}} and {{math|d(x_m, x) < ε/2}}, where {{math|ε > 0}} is given and arbitrary (as in the definition of a limit in a metric space), then by the triangle inequality, {{math|d(x_n, x_m) ≤ d(x_n, x) + d(x_m, x) < ε/2 + ε/2 {{=}} ε}}, so that the sequence {{math|{{mset|x_n}}}} is a Cauchy sequence, by definition.

This version of the triangle inequality reduces to the one stated above in case of normed vector spaces where a metric is induced via {{math|d(u, v) ≔ ‖u − v‖}}, with {{math|u − v}} being the vector pointing from point {{mvar|v}} to {{mvar|u}}.

{{cite book |title=The popular educator; fourth volume |chapter-url=https://books.google.com/books?id=lTACAAAAQAAJ&pg=PA196 |page=196 |chapter=Exercise I. to proposition XIX |year=1854 |publisher=John Cassell |location=Ludgate Hill, London |author=Anonymous}}

:Any side of a triangle is greater than or equal to the difference between the other two sides.

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

: $\backslash big|\backslash |u\backslash |-\backslash |v\backslash |\backslash big|\; \backslash leq\; \backslash |u-v\backslash |,$

or for metric spaces, $|d(A,\; C)\; -\; d(B,\; C)|\; \backslash leq\; d(A,\; B)$.

This implies that the norm $\backslash |\backslash cdot\backslash |$ as well as the distance-from-$z$ function $d(z\; ,\backslash cdot)$ are Lipschitz continuous with Lipschitz constant {{math|1}}, and therefore are in particular uniformly continuous.

The proof of the reverse triangle inequality from the usual one uses $\backslash |v-u\backslash |\; =\; \backslash |\{-\}1(u-v)\backslash |\; =\; |\{-\}1|\backslash cdot\backslash |u-v\backslash |\; =\; \backslash |u-v\backslash |$ to find:

: $\backslash |u\backslash |\; =\; \backslash |(u-v)\; +\; v\backslash |\; \backslash leq\; \backslash |u-v\backslash |\; +\; \backslash |v\backslash |\; \backslash Rightarrow\; \backslash |u\backslash |\; -\; \backslash |v\backslash |\; \backslash leq\; \backslash |u-v\backslash |,$

: $\backslash |v\backslash |\; =\; \backslash |(v-u)\; +\; u\backslash |\; \backslash leq\; \backslash |v-u\backslash |\; +\; \backslash |u\backslash |\; \backslash Rightarrow\; \backslash |u\backslash |\; -\; \backslash |v\backslash |\; \backslash geq\; -\backslash |u-v\backslash |,$

Combining these two statements gives:

: $-\backslash |u-v\backslash |\; \backslash leq\; \backslash |u\backslash |-\backslash |v\backslash |\; \backslash leq\; \backslash |u-v\backslash |\; \backslash Rightarrow\; \backslash big|\backslash |u\backslash |-\backslash |v\backslash |\backslash big|\; \backslash leq\; \backslash |u-v\backslash |.$

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

If $\backslash |u\; +v\backslash |\; -\; \backslash |u\backslash |\; \backslash geq\; 0,$ then by the reverse triangle inequality, $\backslash |u\; +v\backslash |\; -\; \backslash |u\backslash |\; =\; \{\backslash big|\}\backslash |u\; +\; v\backslash |-\backslash |u\backslash |\{\backslash big|\}\; \backslash leq\; \backslash |(u\; +\; v)\; -\; u\backslash |\; =\; \backslash |v\backslash |\; \backslash Rightarrow\; \backslash |u\; +\; v\backslash |\; \backslash leq\; \backslash |u\backslash |\; +\; \backslash |v\backslash |$,

and if then trivially $\backslash |u\backslash |\; +\backslash |v\backslash |\; \backslash geq\; \backslash |u\backslash |\; >\; \backslash |u\; +\; v\backslash |$ by the nonnegativity of the norm.

Thus, in both cases, we find that $\backslash |u\backslash |\; +\; \backslash |v\backslash |\; \backslash geq\; \backslash |u\; +\; v\backslash |$.

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

$d(A,\; B)\; +\; d(B,\; C)\; \backslash geq\; d(A,\; C)\; \backslash Rightarrow\; d(A,\; B)\; \backslash geq\; d(A,\; C)\; -\; d(B,\; C)$

$d(C,\; A)\; +\; d(A,\; B)\; \backslash geq\; d(C,\; B)\; \backslash Rightarrow\; d(A,\; B)\; \backslash geq\; d(B,\; C)\; -\; d(A,\; C)$

Putting these equations together we find:

$d(A,\; B)\; \backslash geq\; |d(A,\; C)\; -\; d(B,\; C)|$

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

If $d(A,\; C)\; -\; d(B,\; C)\; \backslash geq\; 0$, then $d(A,\; B)\; \backslash geq\; |d(A,\; C)\; -\; d(B,\; C)|\; =\; d(A,\; C)\; -\; d(B,\; C)\; \backslash Rightarrow\; d(A,\; B)\; +\; d(B,\; C)\; \backslash geq\; d(A,\; C)$,

and if then $d(A,\; B)\; +\; d(B,\; C)\; \backslash geq\; d(B,\; C)\; >\; d(A,\; C)$ again by the nonnegativity of the metric.

Thus, in both cases, we find that $d(A,\; B)\; +\; d(B,\; C)\; \backslash geq\; d(A,\; C)$.

By applying the cosine function to the triangle inequality and reverse triangle inequality for arc lengths and employing the angle addition and subtraction formulas for cosines, it follows immediately that{{cite conference |last=Schubert |first=Erich |title=A Triangle Inequality for Cosine Similarity |conference=International Conference on Similarity Search and Applications |date=2021 |location=Dortmund |doi=10.1007/978-3-030-89657-7_3 |publisher=Springer |arxiv=2107.04071 }}

$$\backslash operatorname\{sim\}(u,w)\; \backslash geq\; \backslash operatorname\{sim\}(u,v)\; \backslash cdot\; \backslash operatorname\{sim\}(v,w)\; -\; \backslash sqrt\{\backslash left(1-\backslash operatorname\{sim\}(u,v)^2\backslash right)\backslash cdot\backslash left(1-\backslash operatorname\{sim\}(v,w)^2\backslash right)\}$$

and

$$\backslash operatorname\{sim\}(u,w)\; \backslash leq\; \backslash operatorname\{sim\}(u,v)\; \backslash cdot\; \backslash operatorname\{sim\}(v,w)\; +\; \backslash sqrt\{\backslash left(1-\backslash operatorname\{sim\}(u,v)^2\backslash right)\backslash cdot\backslash left(1-\backslash operatorname\{sim\}(v,w)^2\backslash right)\}\backslash ,.$$

With these formulas, one needs to compute a square root for each triple of vectors {{math|{u, v, w}}} that is examined rather than {{math|arccos(sim(u,v))}} for each pair of vectors {{math|{u, v}}} examined, and could be a performance improvement when the number of triples examined is less than the number of pairs examined.

: $\backslash |u+v\backslash |\; \backslash geq\; \backslash |u\backslash |\; +\; \backslash |v\backslash |.$

A physical example of this inequality is the twin paradox in special relativity. The same reversed form of the inequality holds if both vectors lie in the past light cone, and if one or both are null vectors. The result holds in $n\; +\; 1$ dimensions for any $n\; \backslash geq\; 1$. If the plane defined by $u$ and $v$ is space-like (and therefore a Euclidean subspace) then the usual triangle inequality holds.

{{DEFAULTSORT:Triangle Inequality}}

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