{"history":[],"article":{"id":"75af0285-7f1d-4be8-9406-6960aac573be","title":"Triangle inequality","slug":"triangle-inequality","content":":::summary\n:::\n\n# \n[[File:TriangleInequality.svg|thumb|Three examples of the triangle inequality for triangles with sides of lengths x, y, z. The top example shows a case where z is much less than the sum\nx + y\nof the other two sides, and the bottom example shows a case where the side z is only slightly less than\nx + y\n.]]\nIn [mathematics](https://bliptext.com/articles/mathematics), the **[triangle](https://bliptext.com/articles/triangle) inequality** states that for any [triangle](https://bliptext.com/articles/triangle), the sum of the lengths of any two sides must be greater than or equal to the length of the remaining side.\nThis statement permits the inclusion of [degenerate triangles](https://bliptext.com/articles/degeneracy-mathematics), but some authors, especially those writing about elementary geometry, will exclude this possibility, thus leaving out the possibility of equality.\nIf a, b, and c are the lengths of the sides of a triangle then the triangle inequality states that\nwith equality only in the degenerate case of a triangle with zero area.\nIn [Euclidean geometry](https://bliptext.com/articles/euclidean-geometry) and some other geometries, the triangle inequality is a theorem about vectors and vector lengths ([norms](https://bliptext.com/articles/norm-mathematics)):\nwhere the length of the third side has been replaced by the length of the vector sum\nu + v\n. When\nu\nand\nv\nare [real numbers](https://bliptext.com/articles/real-number), they can be viewed as vectors in \\R^1, and the triangle inequality expresses a relationship between [absolute values](https://bliptext.com/articles/absolute-value).\nIn Euclidean geometry, for [right triangles](https://bliptext.com/articles/right-triangle) the triangle inequality is a consequence of the [Pythagorean theorem](https://bliptext.com/articles/pythagorean-theorem), and for general triangles, a consequence of the [law of cosines](https://bliptext.com/articles/law-of-cosines), although it may be proved without these theorems.\nThe inequality can be viewed intuitively in either \\R^2 or \\R^3.\nThe figure at the right shows three examples beginning with clear inequality (top) and approaching equality (bottom).\nIn the Euclidean case, equality occurs only if the triangle has a\n180°\nangle and two\n0°\nangles, making the three [vertices](https://bliptext.com/articles/vertex-geometry) [collinear](https://bliptext.com/articles/straight-line), as shown in the bottom example.\nThus, in Euclidean geometry, the shortest distance between two points is a straight line.\nIn [spherical geometry](https://bliptext.com/articles/spherical-geometry), the shortest distance between two points is an arc of a [great circle](https://bliptext.com/articles/great-circle), but the triangle inequality holds provided the restriction is made that the distance between two points on a sphere is the length of a minor spherical line segment (that is, one with central angle in\n) with those endpoints.\nThe triangle inequality is a defining property of [norms](https://bliptext.com/articles/norm-mathematics) and measures of [distance](https://bliptext.com/articles/metric-mathematics).\nThis property must be established as a theorem for any function proposed for such purposes for each particular space: for example, spaces such as the [real numbers](https://bliptext.com/articles/real-number), [Euclidean spaces](https://bliptext.com/articles/euclidean-space), the [Lp spaces](https://bliptext.com/articles/lp-space) (\np ≥ 1\n), and [inner product spaces](https://bliptext.com/articles/inner-product-space).\n\n# Euclidean geometry\nEuclid proved the triangle inequality for distances in [plane geometry](https://bliptext.com/articles/euclidean-geometry) using the construction in the figure.\nBeginning with triangle ABC, an isosceles triangle is constructed with one side taken as BC and the other equal leg BD along the extension of side AB.\nIt then is argued that angle β has larger measure than angle α, so side AD is longer than side AC.\nHowever:\nso the sum of the lengths of sides AB and BC is larger than the length of AC.\nThis proof appears in [Euclid's Elements](https://bliptext.com/articles/euclid-s-elements), Book 1, Proposition 20.\n# [Euclid's construction for proof of the triangle inequality for plane geometry. | upload.wikimedia.org/wikipedia/commons/8/8d/Euclid///triangle///inequality.svg]\n\n## Mathematical expression of the constraint on the sides of a triangle\nFor a proper triangle, the triangle inequality, as stated in words, literally translates into three inequalities (given that a proper triangle has side lengths\na, b, c\nthat are all positive and excludes the degenerate case of zero area): A more succinct form of this inequality system can be shown to be Another way to state it is implying and thus that the longest side length is less than the [semiperimeter](https://bliptext.com/articles/semiperimeter).\nA mathematically equivalent formulation is that the area of a triangle with sides\na, b, c\nmust be a real number greater than zero.\n[Heron's formula](https://bliptext.com/articles/heron-s-formula) for the area is\nIn 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).\nThe triangle inequality provides two more interesting constraints for triangles whose sides are\na, b, c\n, where\na ≥ b ≥ c\nand \\phi is the [golden ratio](https://bliptext.com/articles/golden-ratio), as\n\n## Right triangle\n[[File:Isosceles triangle made of right triangles.svg|thumb|Isosceles triangle with equal sides\ndivided into two right triangles by an altitude drawn from one of the two base angles.]]\nIn 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.\nThe second part of this theorem is already established above for any side of any triangle.\nThe first part is established using the lower figure.\nIn the figure, consider the right triangle ADC.\nAn isosceles triangle ABC is constructed with equal sides\n. From the [triangle postulate](https://bliptext.com/articles/triangle-postulate), the angles in the right triangle\nADC\nsatisfy:\nLikewise, in the isosceles triangle ABC, the angles satisfy:\nTherefore,\nand so, in particular,\nThat means side AD, which is opposite to angle α, is shorter than side AB, which is opposite to the larger angle β.\nBut\n. Hence:\nA similar construction shows\nAC > DC\n, establishing the theorem.\nAn alternative proof (also based upon the [triangle postulate](https://bliptext.com/articles/triangle-postulate)) proceeds by considering three positions for point B: (i) as depicted (which is to be proved), or (ii) B coincident with D (which would mean the isosceles triangle had two right angles as base angles plus the vertex angle γ, which would violate the [triangle postulate](https://bliptext.com/articles/triangle-postulate)), or lastly, (iii) B interior to the right triangle between points A and D (in which case angle ABC is an exterior angle of a right triangle BDC and therefore larger than\nπ/2\n, meaning the other base angle of the isosceles triangle also is greater than\nπ/2\nand their sum exceeds π in violation of the triangle postulate).\nThis theorem establishing inequalities is sharpened by [Pythagoras' theorem](https://bliptext.com/articles/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.\n\n## Examples of use\nConsider a triangle whose sides are in an [arithmetic progression](https://bliptext.com/articles/arithmetic-progression) and let the sides be\na, a + d, a + 2d\n. Then the triangle inequality requires that\nTo satisfy all these inequalities requires\nWhen d is chosen such that\n, it generates a right triangle that is always similar to the [Pythagorean triple](https://bliptext.com/articles/pythagorean-triple) with sides\n3\n,\n4\n,\n5\n.\nNow consider a triangle whose sides are in a [geometric progression](https://bliptext.com/articles/geometric-progression) and let the sides be\na, ar, ar2\n. Then the triangle inequality requires that\nThe first inequality requires\na > 0\na > 0\n, the middle inequality only requires\nr > 0\n. This now leaves the first and third inequalities needing to satisfy\nThe first of these quadratic inequalities requires r to range in the region beyond the value of the positive root of the quadratic equation\nr2 + r − 1 = 0\n, i.e.\nr > φ − 1\nwhere φ is the [golden ratio](https://bliptext.com/articles/golden-ratio).\nThe second quadratic inequality requires r to range between 0 and the positive root of the quadratic equation\nr2 − r − 1 = 0\n, i.e.\n0 < r < φ\n. The combined requirements result in r being confined to the range\nWhen r the common ratio is chosen such that\nit generates a right triangle that is always similar to the [Kepler triangle](https://bliptext.com/articles/kepler-triangle).\n\n## Generalization to any polygon\nThe triangle inequality can be extended by [mathematical induction](https://bliptext.com/articles/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.\nConsequently, the length of any polygon side is always less than the sum of the other polygon side lengths.\n\n### Example of the generalized polygon inequality for a quadrilateral\nConsider a quadrilateral whose sides are in a [geometric progression](https://bliptext.com/articles/geometric-progression) and let the sides be\na, ar, ar2, ar3\n. Then the generalized polygon inequality requires that\nThese inequalities for\na > 0\nreduce to the following\nThe left-hand side polynomials of these two inequalities have roots that are the [tribonacci constant](https://bliptext.com/articles/generalizations-of-fibonacci-numbers) and its reciprocal.\nConsequently, r is limited to the range\n1/t < r < t\nwhere t is the tribonacci constant.\n\n### Relationship with shortest paths\nThis generalization can be used to prove that the shortest curve between two points in Euclidean geometry is a straight line.\nNo polygonal path between two points is shorter than the line between them.\nThis implies that no curve can have an [arc length](https://bliptext.com/articles/arc-length) less than the distance between its endpoints.\nBy definition, the arc length of a curve is the [least upper bound](https://bliptext.com/articles/least-upper-bound) of the lengths of all polygonal approximations of the curve.\nThe result for polygonal paths shows that the straight line between the endpoints is the shortest of all the polygonal approximations.\nBecause 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.\n# [The arc length of a curve is defined as the least upper bound of the lengths of polygonal approximations. | upload.wikimedia.org/wikipedia/commons/5/5a/Arclength.svg]\n\n## Converse\nThe 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.\nIn either case, if the side lengths are a, b, c we can attempt to place a triangle in the [Euclidean plane](https://bliptext.com/articles/euclidean-plane) as shown in the diagram.\nWe need to prove that there exists a real number h consistent with the values a, b, and c, in which case this triangle exists.\n[[Image:Triangle with notations 3.svg|thumb|270px|Triangle with altitude h cutting base c into\nd + (c − d)\n.]]\nBy the [Pythagorean theorem](https://bliptext.com/articles/pythagorean-theorem) we have\nand\naccording to the figure at the right.\nSubtracting these yields\na2 − b2 = c2 − 2cd\n. This equation allows us to express d in terms of the sides of the triangle:\nFor the height of the triangle we have that\n. By replacing d with the formula given above, we have\nFor a real number h to satisfy this,\nh2\nmust be non-negative:\nwhich holds if the triangle inequality is satisfied for all sides.\nTherefore, there does exist a real number h consistent with the sides a, b, c, and the triangle exists.\nIf each triangle inequality holds [strictly](https://bliptext.com/articles/strict-inequality), 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.\n\n## Generalization to higher dimensions\nThe area of a triangular face of a [tetrahedron](https://bliptext.com/articles/tetrahedron) is less than or equal to the sum of the areas of the other three triangular faces.\nMore generally, in Euclidean space the hypervolume of an\n(n − 1)\n-[facet](https://bliptext.com/articles/facet-mathematics) of an n-[simplex](https://bliptext.com/articles/simplex) is less than or equal to the sum of the hypervolumes of the other n [facet](https://bliptext.com/articles/facet-mathematics)s.\nMuch as the triangle inequality generalizes to a polygon inequality, the inequality for a simplex of any dimension generalizes to a [polytope](https://bliptext.com/articles/polytope) of any dimension: the hypervolume of any facet of a [polytope](https://bliptext.com/articles/polytope) is less than or equal to the sum of the hypervolumes of the remaining facets.\nIn some cases the tetrahedral inequality is stronger than several applications of the triangle inequality.\nFor example, the triangle inequality appears to allow the possibility of four points A, B, C, and Z in Euclidean space such that distances\nAB = BC = CA = 26\nand\nAZ = BZ = CZ = 14\n. However, points with such distances cannot exist: the area of the\n26–26–26\nequilateral triangle\nABC\nis 169\\sqrt 3, which is larger than three times 39\\sqrt 3, the area of a\n26–14–14\nisosceles triangle (all by [Heron's formula](https://bliptext.com/articles/heron-s-formula)), and so the arrangement is forbidden by the tetrahedral inequality.\n\n# Normed vector space\nIn a [normed vector space](https://bliptext.com/articles/[norm](https://bliptext.com/articles/norm-mathematics)ed-vector-space) V, one of the defining properties of the [norm](https://bliptext.com/articles/norm-mathematics) is the triangle inequality:\nThat is, the norm of the [sum of two vectors](https://bliptext.com/articles/vector-sum) is at most as large as the sum of the norms of the two vectors.\nThis is also referred to as [subadditivity](https://bliptext.com/articles/subadditivity).\nFor any proposed function to behave as a norm, it must satisfy this requirement.\nIf the normed space is [Euclidean](https://bliptext.com/articles/euclidean-space), or, more generally, [strictly convex](https://bliptext.com/articles/strictly-convex-space), then if and only if the triangle formed by\nu\n,\nv\n, and\nu + v\n, is degenerate, that is,\nu\nand\nv\nare on the same ray, i.e.,\nor\n, or\nfor some\nα > 0\n. This property characterizes strictly convex normed spaces such as the\nℓp\nspaces with\n1 < p < ∞\n. However, there are normed spaces in which this is not true.\nFor instance, consider the plane with the\nℓ1\nnorm (the [Manhattan distance](https://bliptext.com/articles/manhattan-distance)) and denote\nand\n. Then the triangle formed by\nu\n,\nv\n, and\nu + v\n, is non-degenerate but\n# [Triangle inequality for norms of vectors. | upload.wikimedia.org/wikipedia/commons/5/5a/Vector-triangle-inequality.svg]\n\n## Example norms\nThe [absolute value](https://bliptext.com/articles/absolute-value) is a norm for the [real line](https://bliptext.com/articles/real-line); as required, the [absolute value](https://bliptext.com/articles/absolute-value) satisfies the triangle inequality for any real numbers u and v. If u and v have the same sign or either of them is zero, then If u and v have opposite signs, then [without loss of generality](https://bliptext.com/articles/without-loss-of-generality) assume |u| > |v|.\nThen Combining these cases:\nThe triangle inequality is useful in [mathematical analysis](https://bliptext.com/articles/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.\nThere is also a lower estimate, which can be found using the reverse triangle inequality which states that for any real numbers u and v,\nThe [taxicab norm](https://bliptext.com/articles/taxicab-norm) or 1-norm is one generalization absolute value to higher dimensions.\nTo find the norm of a vector just add the absolute value of each component separately,\nThe Euclidean norm or 2-norm defines the length of translation vectors in an n-dimensional [Euclidean space](https://bliptext.com/articles/euclidean-space) in terms of a [Cartesian coordinate system](https://bliptext.com/articles/cartesian-coordinate-system).\nFor a vector its length is defined using the n-dimensional [Pythagorean theorem](https://bliptext.com/articles/pythagorean-theorem):\nThe inner product is norm in any [inner product space](https://bliptext.com/articles/inner-product-space), a generalization of Euclidean vector spaces including infinite-dimensional examples.\nThe triangle inequality follows from the [Cauchy–Schwarz inequality](https://bliptext.com/articles/cauchy-schwarz-inequality) as follows: Given vectors u and v, and denoting the inner product as :\nThe Cauchy–Schwarz inequality turns into an equality if and only if u and v are linearly dependent.\nThe inequality turns into an equality for linearly dependent u and v if and only if one of the vectors u or v is a nonnegative scalar of the other.\nTaking the square root of the final result gives the triangle inequality.\nThe [p-norm](https://bliptext.com/articles/p-norm) is a generalization of taxicab and Euclidean norms, using an arbitrary positive integer exponent, where the\nvi\nare the components of vector v.\nExcept for the case\np = 2\n, the p-norm is not an inner product norm, because it does not satisfy the [parallelogram law](https://bliptext.com/articles/parallelogram-law).\nThe triangle inequality for general values of p is called [Minkowski's inequality](https://bliptext.com/articles/minkowski-s-inequality).\nIt takes the form:\n\n# Metric space\nIn a [metric space](https://bliptext.com/articles/metric-space) M with metric d, the triangle inequality is a requirement upon [distance](https://bliptext.com/articles/metric-mathematics):\nfor all points A, B, and C in M. That is, the distance from A to C is at most as large as the sum of the distance from A to B and the distance from B to C.\nThe triangle inequality is responsible for most of the interesting structure on a metric space, namely, convergence.\nThis is because the remaining requirements for a metric are rather simplistic in comparison.\nFor example, the fact that any [convergent sequence](https://bliptext.com/articles/limit-of-a-sequence) in a metric space is a [Cauchy sequence](https://bliptext.com/articles/cauchy-sequence) is a direct consequence of the triangle inequality, because if we choose any\nxn\nand\nxm\nsuch that\nd(xn, x) < ε/2\nand\nd(xm, x) < ε/2\n, where\nε > 0\nis given and arbitrary (as in the definition of a limit in a metric space), then by the triangle inequality,\nd(xn, xm) ≤ d(xn, x) + d(xm, x) < ε/2 + ε/2 = ε\n, so that the sequence\nis a Cauchy sequence, by definition.\nThis version of the triangle inequality reduces to the one stated above in case of normed vector spaces where a metric is induced via\nd(u, v) ≔ ‖u − v‖\n, with\nu − v\nbeing the vector pointing from point v to u.\n\n# Reverse triangle inequality\nThe **reverse triangle inequality** is an equivalent alternative formulation of the triangle inequality that gives lower bounds instead of upper bounds.\nFor plane geometry, the statement is:\nIn the case of a normed vector space, the statement is:\nor for metric spaces,.\nThis implies that the norm \\|\\cdot\\| as well as the distance-from-z function d(z ,\\cdot) are [Lipschitz continuous](https://bliptext.com/articles/lipschitz-continuity) with Lipschitz constant\n1\n, and therefore are in particular [uniformly continuous](https://bliptext.com/articles/uniform-continuity).\nThe proof of the reverse triangle inequality from the usual one uses to find:\nCombining these two statements gives:\nIn the converse, the proof of the triangle inequality from the reverse triangle inequality works in two cases:\nIf then by the reverse triangle inequality, ,\nand if then trivially by the nonnegativity of the norm.\nThus, in both cases, we find that.\nFor metric spaces, the proof of the reverse triangle inequality is found similarly by:\nPutting these equations together we find:\nAnd in the converse, beginning from the reverse triangle inequality, we can again use two cases:\nIf, then ,\nand if then again by the nonnegativity of the metric.\nThus, in both cases, we find that.\n\n# Triangle inequality for cosine similarity\nBy 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\nand\nWith these formulas, one needs to compute a [square root](https://bliptext.com/articles/square-root) for each triple of vectors\n{u, v, w}\nthat is examined rather than\narccos(sim(u,v))\nfor each pair of vectors\n{u, v}\nexamined, and could be a performance improvement when the number of triples examined is less than the number of pairs examined.\n\n# Reversal in Minkowski space\nThe [Minkowski space](https://bliptext.com/articles/minkowski-space) metric is not positive-definite, which means that can have either sign or vanish, even if the vector u is non-zero.\nMoreover, if u and v are both timelike vectors lying in the future light cone, the triangle inequality is reversed:\nA physical example of this inequality is the [twin paradox](https://bliptext.com/articles/twin-paradox) in [special relativity](https://bliptext.com/articles/special-relativity).\nThe same reversed form of the inequality holds if both vectors lie in the past light cone, and if one or both are null vectors.\nThe result holds in n + 1 dimensions for any n \\geq 1.\nIf the plane defined by u and v is space-like (and therefore a Euclidean subspace) then the usual triangle inequality holds.","current_revision":null,"created_by":"kACnuQmH5k9Nr5VNVEpVeg7XpvUmMctD","created_at":"2025-01-25T17:38:22.284Z","updated_at":"2025-01-25T17:38:22.284Z","search_vector":"'inequ':2 'triangl':1","search_content":" \n \n\n# \n[[File:TriangleInequality.svg|thumb|Three examples of the triangle inequality for triangles with sides of lengths x, y, z. The top example shows a case where z is much less than the sum\nx + y\nof the other two sides, and the bottom example shows a case where the side z is only slightly less than\nx + y\n.]]\nIn [mathematics](https://bliptext.com/articles/mathematics), the **[triangle](https://bliptext.com/articles/triangle) inequality** states that for any [triangle](https://bliptext.com/articles/triangle), the sum of the lengths of any two sides must be greater than or equal to the length of the remaining side.\nThis statement permits the inclusion of [degenerate triangles](https://bliptext.com/articles/degeneracy-mathematics), but some authors, especially those writing about elementary geometry, will exclude this possibility, thus leaving out the possibility of equality.\nIf a, b, and c are the lengths of the sides of a triangle then the triangle inequality states that\nwith equality only in the degenerate case of a triangle with zero area.\nIn [Euclidean geometry](https://bliptext.com/articles/euclidean-geometry) and some other geometries, the triangle inequality is a theorem about vectors and vector lengths ([norms](https://bliptext.com/articles/norm-mathematics)):\nwhere the length of the third side has been replaced by the length of the vector sum\nu + v\n. When\nu\nand\nv\nare [real numbers](https://bliptext.com/articles/real-number), they can be viewed as vectors in \\R^1, and the triangle inequality expresses a relationship between [absolute values](https://bliptext.com/articles/absolute-value).\nIn Euclidean geometry, for [right triangles](https://bliptext.com/articles/right-triangle) the triangle inequality is a consequence of the [Pythagorean theorem](https://bliptext.com/articles/pythagorean-theorem), and for general triangles, a consequence of the [law of cosines](https://bliptext.com/articles/law-of-cosines), although it may be proved without these theorems.\nThe inequality can be viewed intuitively in either \\R^2 or \\R^3.\nThe figure at the right shows three examples beginning with clear inequality (top) and approaching equality (bottom).\nIn the Euclidean case, equality occurs only if the triangle has a\n180°\nangle and two\n0°\nangles, making the three [vertices](https://bliptext.com/articles/vertex-geometry) [collinear](https://bliptext.com/articles/straight-line), as shown in the bottom example.\nThus, in Euclidean geometry, the shortest distance between two points is a straight line.\nIn [spherical geometry](https://bliptext.com/articles/spherical-geometry), the shortest distance between two points is an arc of a [great circle](https://bliptext.com/articles/great-circle), but the triangle inequality holds provided the restriction is made that the distance between two points on a sphere is the length of a minor spherical line segment (that is, one with central angle in\n) with those endpoints.\nThe triangle inequality is a defining property of [norms](https://bliptext.com/articles/norm-mathematics) and measures of [distance](https://bliptext.com/articles/metric-mathematics).\nThis property must be established as a theorem for any function proposed for such purposes for each particular space: for example, spaces such as the [real numbers](https://bliptext.com/articles/real-number), [Euclidean spaces](https://bliptext.com/articles/euclidean-space), the [Lp spaces](https://bliptext.com/articles/lp-space) (\np ≥ 1\n), and [inner product spaces](https://bliptext.com/articles/inner-product-space).\n\n# Euclidean geometry\nEuclid proved the triangle inequality for distances in [plane geometry](https://bliptext.com/articles/euclidean-geometry) using the construction in the figure.\nBeginning with triangle ABC, an isosceles triangle is constructed with one side taken as BC and the other equal leg BD along the extension of side AB.\nIt then is argued that angle β has larger measure than angle α, so side AD is longer than side AC.\nHowever:\nso the sum of the lengths of sides AB and BC is larger than the length of AC.\nThis proof appears in [Euclid's Elements](https://bliptext.com/articles/euclid-s-elements), Book 1, Proposition 20.\n# [Euclid's construction for proof of the triangle inequality for plane geometry. | upload.wikimedia.org/wikipedia/commons/8/8d/Euclid///triangle///inequality.svg]\n\n## Mathematical expression of the constraint on the sides of a triangle\nFor a proper triangle, the triangle inequality, as stated in words, literally translates into three inequalities (given that a proper triangle has side lengths\na, b, c\nthat are all positive and excludes the degenerate case of zero area): A more succinct form of this inequality system can be shown to be Another way to state it is implying and thus that the longest side length is less than the [semiperimeter](https://bliptext.com/articles/semiperimeter).\nA mathematically equivalent formulation is that the area of a triangle with sides\na, b, c\nmust be a real number greater than zero.\n[Heron's formula](https://bliptext.com/articles/heron-s-formula) for the area is\nIn 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).\nThe triangle inequality provides two more interesting constraints for triangles whose sides are\na, b, c\n, where\na ≥ b ≥ c\nand \\phi is the [golden ratio](https://bliptext.com/articles/golden-ratio), as\n\n## Right triangle\n[[File:Isosceles triangle made of right triangles.svg|thumb|Isosceles triangle with equal sides\ndivided into two right triangles by an altitude drawn from one of the two base angles.]]\nIn 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.\nThe second part of this theorem is already established above for any side of any triangle.\nThe first part is established using the lower figure.\nIn the figure, consider the right triangle ADC.\nAn isosceles triangle ABC is constructed with equal sides\n. From the [triangle postulate](https://bliptext.com/articles/triangle-postulate), the angles in the right triangle\nADC\nsatisfy:\nLikewise, in the isosceles triangle ABC, the angles satisfy:\nTherefore,\nand so, in particular,\nThat means side AD, which is opposite to angle α, is shorter than side AB, which is opposite to the larger angle β.\nBut\n. Hence:\nA similar construction shows\nAC > DC\n, establishing the theorem.\nAn alternative proof (also based upon the [triangle postulate](https://bliptext.com/articles/triangle-postulate)) proceeds by considering three positions for point B: (i) as depicted (which is to be proved), or (ii) B coincident with D (which would mean the isosceles triangle had two right angles as base angles plus the vertex angle γ, which would violate the [triangle postulate](https://bliptext.com/articles/triangle-postulate)), or lastly, (iii) B interior to the right triangle between points A and D (in which case angle ABC is an exterior angle of a right triangle BDC and therefore larger than\nπ/2\n, meaning the other base angle of the isosceles triangle also is greater than\nπ/2\nand their sum exceeds π in violation of the triangle postulate).\nThis theorem establishing inequalities is sharpened by [Pythagoras' theorem](https://bliptext.com/articles/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.\n\n## Examples of use\nConsider a triangle whose sides are in an [arithmetic progression](https://bliptext.com/articles/arithmetic-progression) and let the sides be\na, a + d, a + 2d\n. Then the triangle inequality requires that\nTo satisfy all these inequalities requires\nWhen d is chosen such that\n, it generates a right triangle that is always similar to the [Pythagorean triple](https://bliptext.com/articles/pythagorean-triple) with sides\n3\n,\n4\n,\n5\n.\nNow consider a triangle whose sides are in a [geometric progression](https://bliptext.com/articles/geometric-progression) and let the sides be\na, ar, ar2\n. Then the triangle inequality requires that\nThe first inequality requires\na > 0\na > 0\n, the middle inequality only requires\nr > 0\n. This now leaves the first and third inequalities needing to satisfy\nThe first of these quadratic inequalities requires r to range in the region beyond the value of the positive root of the quadratic equation\nr2 + r − 1 = 0\n, i.e.\nr > φ − 1\nwhere φ is the [golden ratio](https://bliptext.com/articles/golden-ratio).\nThe second quadratic inequality requires r to range between 0 and the positive root of the quadratic equation\nr2 − r − 1 = 0\n, i.e.\n0 0\nreduce to the following\nThe left-hand side polynomials of these two inequalities have roots that are the [tribonacci constant](https://bliptext.com/articles/generalizations-of-fibonacci-numbers) and its reciprocal.\nConsequently, r is limited to the range\n1/t 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.\n\n## Generalization to higher dimensions\nThe area of a triangular face of a [tetrahedron](https://bliptext.com/articles/tetrahedron) is less than or equal to the sum of the areas of the other three triangular faces.\nMore generally, in Euclidean space the hypervolume of an\n(n − 1)\n-[facet](https://bliptext.com/articles/facet-mathematics) of an n-[simplex](https://bliptext.com/articles/simplex) is less than or equal to the sum of the hypervolumes of the other n [facet](https://bliptext.com/articles/facet-mathematics)s.\nMuch as the triangle inequality generalizes to a polygon inequality, the inequality for a simplex of any dimension generalizes to a [polytope](https://bliptext.com/articles/polytope) of any dimension: the hypervolume of any facet of a [polytope](https://bliptext.com/articles/polytope) is less than or equal to the sum of the hypervolumes of the remaining facets.\nIn some cases the tetrahedral inequality is stronger than several applications of the triangle inequality.\nFor example, the triangle inequality appears to allow the possibility of four points A, B, C, and Z in Euclidean space such that distances\nAB = BC = CA = 26\nand\nAZ = BZ = CZ = 14\n. However, points with such distances cannot exist: the area of the\n26–26–26\nequilateral triangle\nABC\nis 169\\sqrt 3, which is larger than three times 39\\sqrt 3, the area of a\n26–14–14\nisosceles triangle (all by [Heron's formula](https://bliptext.com/articles/heron-s-formula)), and so the arrangement is forbidden by the tetrahedral inequality.\n\n# Normed vector space\nIn a [[norm](https://bliptext.com/articles/norm-mathematics)ed vector space](https://bliptext.com/articles/[norm](https://bliptext.com/articles/norm-mathematics)ed-vector-space) V, one of the defining properties of the [norm](https://bliptext.com/articles/norm-mathematics) is the triangle inequality:\nThat is, the norm of the [sum of two vectors](https://bliptext.com/articles/vector-sum) is at most as large as the sum of the norms of the two vectors.\nThis is also referred to as [subadditivity](https://bliptext.com/articles/subadditivity).\nFor any proposed function to behave as a norm, it must satisfy this requirement.\nIf the normed space is [Euclidean](https://bliptext.com/articles/euclidean-space), or, more generally, [strictly convex](https://bliptext.com/articles/strictly-convex-space), then if and only if the triangle formed by\nu\n,\nv\n, and\nu + v\n, is degenerate, that is,\nu\nand\nv\nare on the same ray, i.e.,\nor\n, or\nfor some\nα > 0\n. This property characterizes strictly convex normed spaces such as the\nℓp\nspaces with\n1 |v|.\nThen Combining these cases:\nThe triangle inequality is useful in [mathematical analysis](https://bliptext.com/articles/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.\nThere is also a lower estimate, which can be found using the reverse triangle inequality which states that for any real numbers u and v,\nThe [taxicab norm](https://bliptext.com/articles/taxicab-norm) or 1-norm is one generalization absolute value to higher dimensions.\nTo find the norm of a vector just add the absolute value of each component separately,\nThe Euclidean norm or 2-norm defines the length of translation vectors in an n-dimensional [Euclidean space](https://bliptext.com/articles/euclidean-space) in terms of a [Cartesian coordinate system](https://bliptext.com/articles/cartesian-coordinate-system).\nFor a vector its length is defined using the n-dimensional [Pythagorean theorem](https://bliptext.com/articles/pythagorean-theorem):\nThe inner product is norm in any [inner product space](https://bliptext.com/articles/inner-product-space), a generalization of Euclidean vector spaces including infinite-dimensional examples.\nThe triangle inequality follows from the [Cauchy–Schwarz inequality](https://bliptext.com/articles/cauchy-schwarz-inequality) as follows: Given vectors u and v, and denoting the inner product as :\nThe Cauchy–Schwarz inequality turns into an equality if and only if u and v are linearly dependent.\nThe inequality turns into an equality for linearly dependent u and v if and only if one of the vectors u or v is a nonnegative scalar of the other.\nTaking the square root of the final result gives the triangle inequality.\nThe [p-norm](https://bliptext.com/articles/p-norm) is a generalization of taxicab and Euclidean norms, using an arbitrary positive integer exponent, where the\nvi\nare the components of vector v.\nExcept for the case\np = 2\n, the p-norm is not an inner product norm, because it does not satisfy the [parallelogram law](https://bliptext.com/articles/parallelogram-law).\nThe triangle inequality for general values of p is called [Minkowski's inequality](https://bliptext.com/articles/minkowski-s-inequality).\nIt takes the form:\n\n# Metric space\nIn a [metric space](https://bliptext.com/articles/metric-space) M with metric d, the triangle inequality is a requirement upon [distance](https://bliptext.com/articles/metric-mathematics):\nfor all points A, B, and C in M. That is, the distance from A to C is at most as large as the sum of the distance from A to B and the distance from B to C.\nThe triangle inequality is responsible for most of the interesting structure on a metric space, namely, convergence.\nThis is because the remaining requirements for a metric are rather simplistic in comparison.\nFor example, the fact that any [convergent sequence](https://bliptext.com/articles/limit-of-a-sequence) in a metric space is a [Cauchy sequence](https://bliptext.com/articles/cauchy-sequence) is a direct consequence of the triangle inequality, because if we choose any\nxn\nand\nxm\nsuch that\nd(xn, x) 0\nis given and arbitrary (as in the definition of a limit in a metric space), then by the triangle inequality,\nd(xn, xm) ≤ d(xn, x) + d(xm, x) < ε/2 + ε/2 = ε\n, so that the sequence\nis a Cauchy sequence, by definition.\nThis version of the triangle inequality reduces to the one stated above in case of normed vector spaces where a metric is induced via\nd(u, v) ≔ ‖u − v‖\n, with\nu − v\nbeing the vector pointing from point v to u.\n\n# Reverse triangle inequality\nThe **reverse triangle inequality** is an equivalent alternative formulation of the triangle inequality that gives lower bounds instead of upper bounds.\nFor plane geometry, the statement is:\nIn the case of a normed vector space, the statement is:\nor for metric spaces,.\nThis implies that the norm \\|\\cdot\\| as well as the distance-from-z function d(z ,\\cdot) are [Lipschitz continuous](https://bliptext.com/articles/lipschitz-continuity) with Lipschitz constant\n1\n, and therefore are in particular [uniformly continuous](https://bliptext.com/articles/uniform-continuity).\nThe proof of the reverse triangle inequality from the usual one uses to find:\nCombining these two statements gives:\nIn the converse, the proof of the triangle inequality from the reverse triangle inequality works in two cases:\nIf then by the reverse triangle inequality, ,\nand if then trivially by the nonnegativity of the norm.\nThus, in both cases, we find that.\nFor metric spaces, the proof of the reverse triangle inequality is found similarly by:\nPutting these equations together we find:\nAnd in the converse, beginning from the reverse triangle inequality, we can again use two cases:\nIf, then ,\nand if then again by the nonnegativity of the metric.\nThus, in both cases, we find that.\n\n# Triangle inequality for cosine similarity\nBy 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\nand\nWith these formulas, one needs to compute a [square root](https://bliptext.com/articles/square-root) for each triple of vectors\n{u, v, w}\nthat is examined rather than\narccos(sim(u,v))\nfor each pair of vectors\n{u, v}\nexamined, and could be a performance improvement when the number of triples examined is less than the number of pairs examined.\n\n# Reversal in Minkowski space\nThe [Minkowski space](https://bliptext.com/articles/minkowski-space) metric is not positive-definite, which means that can have either sign or vanish, even if the vector u is non-zero.\nMoreover, if u and v are both timelike vectors lying in the future light cone, the triangle inequality is reversed:\nA physical example of this inequality is the [twin paradox](https://bliptext.com/articles/twin-paradox) in [special relativity](https://bliptext.com/articles/special-relativity).\nThe same reversed form of the inequality holds if both vectors lie in the past light cone, and if one or both are null vectors.\nThe result holds in n + 1 dimensions for any n \\geq 1.\nIf the plane defined by u and v is space-like (and therefore a Euclidean subspace) then the usual triangle inequality holds.","revisionCount":0},"hasMore":false,"nextCursor":null}

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