Editing Triangle inequality (section)

==Euclidean geometry==

[[File:Euclid triangle inequality.svg|thumb|Euclid's construction for proof of the triangle inequality for plane geometry.]]

The triangle inequality theorem is stated in [[Euclid's Elements]], Book 1, Proposition 20:

''[…] in the triangle ABC the sum of any two sides is greater than the remaining one, that is, the sum of BA and AC is greater than BC, the sum of AB and BC is greater than AC, and the sum of BC and CA is greater than AB.'' <ref>(trans. Joyce, see reference below)</ref>

Euclid proved the triangle inequality for distances in [[Euclidean geometry|plane geometry]] using the construction in the figure.<ref name=Jacobs>
{{cite book |page=201 |author=Harold R. Jacobs |title=Geometry: seeing, doing, understanding |url=https://books.google.com/books?id=XhQRgZRDDq0C&pg=PA201 |isbn=0-7167-4361-2 |edition=3rd |publisher=Macmillan |year=2003}}
</ref> Beginning with triangle {{mvar|ABC}}, an isosceles triangle is constructed with one side taken as {{mvar|{{overline|BC}}}} and the other equal leg {{mvar|{{overline|BD}}}} along the extension of side {{mvar|{{overline|AB}}}}. It then is argued that angle {{mvar|β}} has larger measure than angle {{mvar|α}}, so side {{mvar|{{overline|AD}}}} is longer than side {{mvar|{{overline|AC}}}}. However: 
:<math>\overline{AD} = \overline{AB} + \overline{BD} = \overline{AB} + \overline{BC},</math> 
so the sum of the lengths of sides {{mvar|{{overline|AB}}}} and {{mvar|{{overline|BC}}}} is larger than the length of {{mvar|{{overline|AC}}}}. This proof appears in [[Euclid's Elements]], Book 1, Proposition 20.<ref name=Joyce>
{{cite web
| url         = http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI20.html
| title       = Euclid's elements, Book 1, Proposition 20
| author      = David E. Joyce | author-link = David E. Joyce (mathematician)
| year        = 1997
| work        = Euclid's elements
| publisher   = Dept. Math and Computer Science, Clark University
| access-date  = 2010-06-25
}}
</ref>

===Mathematical expression of the constraint on the sides of a triangle===

For a proper triangle, the triangle inequality, as stated in words, literally translates into three inequalities (given that a proper triangle has side lengths {{math|''a'', ''b'', ''c''}} that are all positive and excludes the degenerate case of zero area):
:<math>a + b > c ,\quad b + c > a ,\quad c + a > b .</math>
A more succinct form of this inequality system can be shown to be
:<math>|a - b| < c < a + b .</math>
Another way to state it is
:<math>\max(a, b, c) < a + b + c - \max(a, b, c)</math>
implying
:<math>2 \max(a, b, c) < a + b + c</math>
and thus that the longest side length is less than the [[semiperimeter]].

A mathematically equivalent formulation is that the area of a triangle with sides {{math|''a'', ''b'', ''c''}} must be a real number greater than zero. [[Heron's formula]] for the area is

:<math>
\begin{align}
4\cdot \text{area} & =\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)} \\
& = \sqrt{-a^4-b^4-c^4+2a^2b^2+2a^2c^2+2b^2c^2}.
\end{align}
</math>

In terms of either area expression, the triangle inequality imposed on all sides is equivalent to the condition that the expression under the square root sign be real and greater than zero (so the area expression is real and greater than zero).

The triangle inequality provides two more interesting constraints for triangles whose sides are {{math|''a'', ''b'', ''c''}}, where {{math|''a'' ≥ ''b'' ≥ ''c''}} and <math>\phi</math> is the [[golden ratio]], as
:<math>1<\frac{a+c}{b}<3</math>

:<math>1\le\min\left(\frac{a}{b}, \frac{b}{c}\right)<\phi.</math><ref>''[[American Mathematical Monthly]]'', pp. 49-50, 1954.</ref>

===Right triangle===

[[File:Isosceles triangle made of right triangles.svg|thumb|Isosceles triangle with equal sides {{math|{{overline|AB}} {{=}} {{overline|AC}}}} divided into two right triangles by an altitude drawn from one of the two base angles.]]

In the case of right triangles, the triangle inequality specializes to the statement that the hypotenuse is greater than either of the two sides and less than their sum.<ref name=Palmer>
{{cite book |title=Practical mathematics for home study: being the essentials of arithmetic, geometry, algebra and trigonometry |author=Claude Irwin Palmer |url=https://archive.org/details/practicalmathema00palmiala |page=[https://archive.org/details/practicalmathema00palmiala/page/422 422] |publisher=McGraw-Hill |year=1919}}
</ref>

The second part of this theorem is already established above for any side of any triangle. The first part is established using the lower figure. In the figure, consider the right triangle {{mvar|ADC}}. An isosceles triangle {{mvar|ABC}} is constructed with equal sides {{math|''{{overline|AB}}'' {{=}} ''{{overline|AC}}''}}. From the [[triangle postulate]], the angles in the right triangle {{math|ADC}} satisfy:
:<math> \alpha + \gamma = \pi /2 \ . </math>
Likewise, in the isosceles triangle {{mvar|ABC}}, the angles satisfy:
:<math>2\beta + \gamma = \pi \ . </math>
Therefore,
:<math> \alpha = \pi/2 - \gamma ,\ \mathrm{while} \ \beta= \pi/2 - \gamma /2  \ ,</math>
and so, in particular,
:<math>\alpha < \beta \ . </math>
That means side {{mvar|AD}}, which is opposite to angle {{mvar|α}}, is shorter than side {{mvar|AB}}, which is opposite to the larger angle {{mvar|β}}. But {{math|''{{overline|AB}}'' {{=}} ''{{overline|AC}}''}}. Hence:
:<math>\overline{AC} > \overline{AD} \ . </math>
A similar construction shows {{math|''{{overline|AC}}'' > ''{{overline|DC}}''}}, establishing the theorem.

An alternative proof (also based upon the triangle postulate) proceeds by considering three positions for point {{mvar|B}}:<ref name=Zawaira>
{{cite book |title=A primer for mathematics competitions |chapter-url=https://books.google.com/books?id=A21T73sqZ3AC&pg=PA30 |chapter=Lemma 1: In a right-angled triangle the hypotenuse is greater than either of the other two sides |author1=Alexander Zawaira |author2=Gavin Hitchcock |isbn=978-0-19-953988-8 |year=2009 |publisher=Oxford University Press}}
</ref> (i) as depicted (which is to be proved), or (ii) {{mvar|B}} coincident with {{mvar|D}} (which would mean the isosceles triangle had two right angles as base angles plus the vertex angle {{mvar|γ}}, which would violate the [[triangle postulate]]), or lastly, (iii) {{mvar|B}} interior to the right triangle between points {{mvar|A}} and {{mvar|D}} (in which case angle {{mvar|ABC}} is an exterior angle of a right triangle {{mvar|BDC}} and therefore larger than {{math|''π''/2}}, meaning the other base angle of the isosceles triangle also is greater than {{math|''π''/2}} and their sum exceeds {{mvar|π}} in violation of the triangle postulate).

This theorem establishing inequalities is sharpened by [[Pythagoras' theorem]] to the equality that the square of the length of the hypotenuse equals the sum of the squares of the other two sides.

===Examples of use===
Consider a triangle whose sides are in an [[arithmetic progression]] and let the sides be {{math|''a'', ''a'' + ''d'', ''a'' + 2''d''}}. Then the triangle inequality requires that

:<math>\begin{array}{rcccl} 
  0 &<& a &<& 2a+3d, \\
  0 &<& a+d &<& 2a+2d, \\
  0 &<& a+2d &<& 2a+d. 
\end{array}</math>

To satisfy all these inequalities requires

:<math> a>0 \text{ and } -\frac{a}{3}<d<a. </math><ref>{{cite journal|title=input: ''solve 0<a<2a+3d, 0<a+d<2a+2d, 0<a+2d<2a+d,'' |last=Wolfram{{!}}Alpha|journal=Wolfram Research|url=http://www.wolframalpha.com/input/?i=solve%200%3Ca%3C2a%2B3d%2C%200%3Ca%2Bd%3C2a%2B2d%2C%200%3Ca%2B2d%3C2a%2Bd&t=ff3tb01|access-date=2010-09-07}}</ref>

When {{mvar|d}} is chosen such that {{math|''d'' {{=}} ''a''/3}}, it generates a right triangle that is always similar to the [[Pythagorean triple]] with sides {{math|3}}, {{math|4}}, {{math|5}}.

Now consider a triangle whose sides are in a [[geometric progression]] and let the sides be {{math|''a'', ''ar'', ''ar''<sup>2</sup>}}. Then the triangle inequality requires that

:<math>\begin{array}{rcccl} 
  0 &<& a &<& ar+ar^2, \\
  0 &<& ar &<& a+ar^2, \\
  0 &<& \! ar^2 &<& a+ar. 
\end{array}</math>

The first inequality requires {{math|''a'' > 0}}; consequently it can be divided through and eliminated. With {{math|''a'' > 0}}, the middle inequality only requires {{math|''r'' > 0}}. This now leaves the first and third inequalities needing to satisfy

:<math>
\begin{align}
r^2+r-1 & {} >0 \\
r^2-r-1 & {} <0.
\end{align}
</math>

The first of these quadratic inequalities requires {{mvar|r}} to range in the region beyond the value of the positive root of the quadratic equation {{math|''r''<sup>2</sup> + ''r'' − 1 {{=}} 0}}, i.e. {{math|''r'' > ''φ'' − 1}}  where {{mvar|φ}} is the [[golden ratio]]. The second quadratic inequality requires {{mvar|r}} to range between 0 and the positive root of the quadratic equation {{math|''r''<sup>2</sup> − ''r'' − 1 {{=}} 0}}, i.e. {{math|0 < ''r'' < ''φ''}}. The combined requirements result in {{mvar|r}} being confined to the range
:<math>\varphi - 1 < r <\varphi\, \text{ and } a >0.</math><ref>{{cite journal|title=input: ''solve 0<a<ar+ar<sup>2</sup>, 0<ar<a+ar<sup>2</sup>, 0<ar<sup>2</sup><a+ar'' |last=Wolfram{{!}}Alpha|journal=Wolfram Research|url=http://wolframalpha.com/input?i=solve+0%3Ca%3Car%2Bar^2%2C+0%3Car%3Ca%2Bar^2%2C+0%3Car^2%3Ca%2Bar|access-date=2010-09-07}}</ref>

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]].

===Generalization to any polygon===
The triangle inequality can be extended by [[mathematical induction]] to arbitrary polygonal paths, showing that the total length of such a path is no less than the length of the straight line between its endpoints. Consequently, the length of any polygon side is always less than the sum of the other polygon side lengths.

====Example of the generalized polygon inequality for a quadrilateral====
Consider a quadrilateral whose sides are in a [[geometric progression]] and let the sides be {{math|''a'', ''ar'', ''ar''<sup>2</sup>, ''ar''<sup>3</sup>}}. Then the generalized polygon inequality requires that

:<math>\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}</math>

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

:<math> r^3+r^2+r-1>0 </math>
:<math> r^3-r^2-r-1<0. </math><ref>{{cite journal|title=input: ''solve 0<a<ar+ar<sup>2</sup>+ar<sup>3</sup>, 0<ar<sup>3</sup><a+ar+ar<sup>2</sup>'' |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}}</ref>
The left-hand side polynomials of these two inequalities have roots that are the [[Generalizations of Fibonacci numbers#Tribonacci numbers|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.

====Relationship with shortest paths====
[[File:Arclength.svg|300px|thumb|The arc length of a curve is defined as the least upper bound of the lengths of polygonal approximations.]]
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.<ref>{{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.</ref>

===Converse===

The converse of the triangle inequality theorem is also true: if three real numbers are such that each is less than the sum of the others, then there exists a triangle with these numbers as its side lengths and with positive area; and if one number equals the sum of the other two, there exists a degenerate triangle (that is, with zero area) with these numbers as its side lengths.

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|thumb|270px|Triangle with altitude {{mvar|h}} cutting base {{mvar|c}} into {{math|''d'' + (''c'' − ''d'')}}.]]

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}} − 2''cd''}}. This equation allows us to express {{mvar|d}} in terms of the sides of the triangle:
:<math>d=\frac{-a^2+b^2+c^2}{2c}.</math>
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

:<math>h^2 = b^2-\left(\frac{-a^2+b^2+c^2}{2c}\right)^2.</math>

For a real number {{mvar|h}} to satisfy this, {{math|''h''{{sup|2}}}} must be non-negative:
:<math>\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}</math>
which holds if the triangle inequality is satisfied for all sides. Therefore, there does exist a real number <math>h</math> consistent with the sides <math>a, b, c</math>, and the triangle exists. If each triangle inequality holds [[strict inequality|strictly]], <math>h > 0</math> and the triangle is non-degenerate (has positive area); but if one of the inequalities holds with equality, so <math>h = 0</math>, the triangle is degenerate.

===Generalization to higher dimensions===
{{unreferenced section|date=October 2021}}

{{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 (mathematics)|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 <math display=inline>169\sqrt 3</math>, which is larger than three times <math display=inline>39\sqrt 3</math>, 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.