Triangle Inequality

In mathematics , the triangle inequality states that for any triangle , the sum of the lengths of any two sides must be greater than or equal to the length of the remaining side. [1] [2] This statement allows for the inclusion of degenerate triangles , but some authors, especially those writing about elementary geometry, would exclude this possibility, thus discarding the possibility of equivalence. [3] If x , y , and z are the lengths of the sides of a triangle with none of the sides greater than z , then the triangle inequality states that

z\leq x+y,

With equality only in the degenerate case of a triangle with zero area. In Euclidean geometry and some other geometry, there is a theorem about triangle inequality distances, and it is written using vectors and vector lengths ( normals ):

Triangle Inequality
Three examples of the triangle inequality for triangles of length x , y , z . The top example shows a case where z is much less than the sum of the other two sides x + y , and the bottom example shows a case where side z is only slightly less than x + y .
\|\mathbf {x} +\mathbf {y} \|\leq \|\mathbf {x} \|+\|\mathbf {y} \|,

where the length z of the third side is replaced by the vector sum x + y . When x and y are real numbers , they can be viewed as vectors in r1 , and the triangle inequality expresses a relationship between net values .

In Euclidean geometry, the triangle inequality for right triangles is a consequence of the Pythagorean theorem , and for normal triangles, is a consequence of the law of cosines , although it can be proved without these theorems. The inequality can be seen intuitively in either R2 or R3 . The figure on the right shows three examples that begin with explicit inequality (top) and near equality (bottom). In the Euclidean case, equivalence is a triangle only if one 180° angle and two 0° angles, making the three vertices collinear . , as shown in the example below. Thus, in Euclidean geometry, the shortest distance between two points is a straight line.

In spherical geometry , the shortest distance between two points is an arc of a great circle , but the triangle inequality condition is that the distance between two points on a sphere is the length of a small circular line segment (that is is provided, with the central angle in a [0, ] ) with those endpoints .

The triangle inequality is a defining property of distance norms and measures . This property must be established as a theorem for any function proposed for such purposes for each particular space: for example, the real numbers , Euclidean spaces , LP spaces ( P 1 ) , and spaces such as inner product spaces .

Euclidean geometry

Triangle Inequality
Euclid’s formulation for the proof of the triangle inequality for plane geometry.

Euclid proved the triangle inequality for distances in plane geometry using the construction in the figure . [6] Starting from the triangle ABC , an isosceles triangle is constructed in which one side is BC and the other equal side BD is along the extension of side AB . Then it is argued that angle β > α , hence side AD > AC . But AD = AB + BD = AB + BC So sum of sides AB + BC> is AC . This proof appears in Euclid’s Elements , Book 1, Proposition 20.

Mathematical expression for the constraints on the sides of a triangle

For a proper triangle, the triangle inequality, as stated in words, literally translates into three inequalities (given that the lengths of the sides of a proper triangle are a , b , c which are all positive and except in the degenerate case of zero field):

a+b>c,\quad b+c>a,\quad c+a>b.

A more concise form of this inequality system can be shown as


another way to tell it

max(a, b, c) < a + b +c -max(a, b, c)

which meant

2max(a, b, c) < a + b + c

And thus the length of the longest side is less than the semiperimeter .

The mathematically equivalent formulation is that the area of ​​a triangle must be a real number with sides a , b , c that are greater than zero. Heron’s formula for area is

{\displaystyle {\begin{aligned}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^{2}b^{2}+2a^{2}c^{2}+2b^{2}c^{2}}}.\end{aligned}}}

In the context of any area expression, the triangle inequality applied on all sides is equal to the condition that the expression under the square root sign is 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 a, b , c , where a b c and is the golden ratio , as

{\displaystyle 1<{\frac {a+c}{b}}<3}
{\displaystyle 1\leq \min \left({\frac {a}{b}},{\frac {b}{c}}\right)<\phi .}

Right triangle

In the case of right-angled triangles, the triangle inequality is characterized by the statement that the hypotenuse is greater than either of the two sides and less than their sum. [9]

The second part of this theorem has already been established above for any side of any triangle. The first part is installed using the lower figure. In the figure, consider a right angled triangle ADC . An isosceles triangle ABC is constructed with equal sides AB = AC . From the triangle concept , the angle ADC in a right triangle satisfies:

Triangle Inequality
An isosceles triangle of equal sides AB = AC is divided into two right-angled triangles by the altitude drawn from one of the two base angles.
\alpha +\gamma =\pi /2\ .

Similarly, in isosceles triangle ABC , angles satisfy:

2\beta +\gamma =\pi \ .


\alpha =\pi /2-\gamma ,\ \mathrm {while} \ \beta =\pi /2-\gamma /2\ ,

and so, in particular,

\alpha <\beta \ .

It means that the angle α opposite the side AD is smaller than the side AB opposite the greater angle β . But AB = AC . That’s why:

{\overline {\mathrm {AC} }}>{\overline {\mathrm {AD} }}\ .

Establishing a similar construction theorem, denotes AC > DC .

An alternative proof (also based on the triangle postulate) proceeds by considering three situations for point B : 10 as shown (which has to be proved), or (ii) B , for D. The triangle had two right angles as the base angle plus the vertex angle coinciding (meaning isosceles) with , which would violate the triangle concept ), or finally, (iii) B for right triangles between the interior points A and D (in which case angle ABC is an exterior angle of a right angled triangle BDC and hence pi /2, which means that the other base angles of the isosceles triangle are also greater than pi/2 and their sum is greater than , in violation of the triangle postulate . This theorem that establishes inequalities is accelerated by Pythagoras’ theorem for the equality that the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides.

Usage examples

Consider a triangle whose sides are in an parallel series and let the sides be a , a + d , a + 2 d . Then the triangle inequality requires that


It is necessary to satisfy all these inequalities

a > 0 and – a/3 < d <a

When d is chosen such that d = a /3 , it forms a right angled triangle that is always similar to a Pythagorean triple whose sides are 3 , 4 , 5 . Now consider a triangle whose sides are in a geometric progression and let the sides be a , ar , ar 2 . Then the triangle inequality requires that

0< a < ar + ar2

0< ar < a + ar2

0< ar2 < a + ar

The first inequality requires a > 0 ; Consequently it can be divided and terminated. With a > 0 , the middle inequality only requires r > 0 . This now leaves the first and third inequalities necessary to satisfy

{\displaystyle {\begin{aligned}r^{2}+r-1&{}>0\\r^{2}-r-1&{}<0.\end{aligned}}}

The first requirement of these quadratic inequalities is r for the series in the region beyond the value of the positive root of the quadratic equation 2 + r – 1 = 0 , i.e. r > – 1 where is the golden ratio. The second quadratic inequality requires the positive root of the quadratic equation of r to range between 0 and 2 – r – 1 = 0 , i.e. 0 < r < . The combined requirements result in r being limited to

{\displaystyle \varphi -1<r<\varphi \,{\text{ and }}a>0.}

When r is the common ratio chosen such that  r = √φ a right angle that is always similar to the Kepler triangle .

Generalization of 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 not less than the length of the straight line between its endpoints. Consequently, the length of any one polygonal side is always less than the sum of the lengths of the other polygonal side.

Example of a generalized polygon inequality for a quadrilateral

Consider a quadrilateral whose sides are in a geometric progression and let the sides be a , ar , ar 2 , ar 3 . Then the generalized polygon inequality requires that

{\displaystyle 0<a<ar+ar^{2}+ar^{3}}

{\displaystyle 0<ar<a+ar^{2}+ar^{3}}

{\displaystyle 0<ar^{2}<a+ar+ar^{3}}

{\displaystyle 0<ar^{3}<a+ar+ar^{2}.}

For a > 0 these inequalities are reduced to

{\displaystyle r^{3}+r^{2}+r-1>0}
{\displaystyle r^{3}-r^{2}-r-1<0.}

The left-hand side of these two inequalities are the roots of the polynomial which are the Tribonacci constant and its reciprocal. Consequently, the r range is limited to 1/ t < r < t where t is the Tribonacci constant.

Shortest path relationship

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

No polygon path between two points is shorter than the line between them. This means that the arc length of any curve cannot be less than the distance between its endpoints. By definition, the arc length of a curve is the lowest upper bound of the length of all polygonal approximations of the curve. The polygon path result shows that the straight line between the end points is the shortest of all polygon projections. Because the arc length of the curve is greater than or equal to the length of each polygonal approximation, the curve itself cannot be shorter than the straight line path. [14]


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 other, then there exists a triangle with these numbers having side lengths and positive area; And if one number is equal to the sum of the other two, then there exists a degenerate triangle (i.e. with zero area) whose lengths of sides coincide with these numbers.

In any case, if the lengths of the sides are a, b, c then we can try to place a triangle in the Euclidean plane as shown in the figure. We need to prove that there exists a real number h corresponding to the values ​​a, b, and c , in which case this triangle exists.

By the Pythagorean theorem we have 2 = 2 + 2 and 2 = 2 (+ c – d ) 2 as in the picture right. Subtracting these gives 2 – 2 = 2 – 2 cd . This equation allows us to express d in terms of the sides of the triangle :

Triangle Inequality
A triangle of height h cuts the base c at d + ( c – d ) .

For the height of the triangle we have 2 = 2 – 2 . Substituting d with the above formula , we have

{\displaystyle h^{2}=b^{2}-\left({\frac {-a^{2}+b^{2}+c^{2}}{2c}}\right)^{2}.}

In order for the real number h to satisfy this, it must be non-negative: h2

{\displaystyle b^{2}-\left({\frac {-a^{2}+b^{2}+c^{2}}{2c}}\right)^{2}\geq 0,}
{\displaystyle \left(b-{\frac {-a^{2}+b^{2}+c^{2}}{2c}}\right)\left(b+{\frac {-a^{2}+b^{2}+c^{2}}{2c}}\right)\geq 0,}
{\displaystyle \left(a^{2}-(b-c)^{2})((b+c)^{2}-a^{2}\right)\geq 0,}
{\displaystyle (a+b-c)(a-b+c)(b+c+a)(b+c-a)\geq 0,}
{\displaystyle (a+b-c)(a+c-b)(b+c-a)\geq 0,}

which holds if the triangle inequality is satisfied for all sides. Therefore there exists a real number h which corresponds to the sides a, b, c and the triangle. If each triangle inequality strictly holds, h > 0 and the triangle is non-degenerative (has a positive area); But if one of the inequalities is accompanied by equality, h = 0, the triangle is degenerate.

Generalizing to Higher Dimensions

In Euclidean space, the hypervolume of a ( n – 1) – aspect of an n – simplex is less than or equal to the sum of the hypervolumes of the other n facets. Specifically, the area of ​​the triangular face of a tetrahedron is less than or equal to the sum of the areas of the other three sides.

Normal vector space

Triangle inequality for normals of vectors.

In an ideal vector space V, one of the defining properties of the ideal is the triangle inequality:

{\displaystyle \|x+y\|\leq \|x\|+\|y\|\quad \forall \,x,y\in V}

That is, the value of the sum of two vectors is at most as large as the sum of the parameters of the two vectors. It is also called subredditivity. For any proposed work to be treated as a criterion, it must meet this requirement. [15] If the ideal space is Euclidean, or, more generally, strictly convex, then if and only if the triangle formed by x , y , and x + y is degenerate, that is, x and y lie on the same ray, that is, x = 0 or y = 0 , or x = α y for some α > 0 ,


This property is characteristic of strictly convex ideal spaces such as p spaces with 1 < p < . However, there are standard places in which this is not true. For example, consider the plane with 1 ideal ( Manhattan distance) AND reflecting x = (1, 0) and y = (0, 1) . So the triangle formed by x , y , and x + y is non-degenerate but


Example criteria

  • Absolute values ​​as ideal for the real line. To be a norm, the triangle inequality requires that the absolute value be satisfied for any real numbers x and y :
|x+y|\leq |x|+|y|,

who does this.


{\displaystyle -\left\vert x\right\vert \leq x\leq \left\vert x\right\vert }
{\displaystyle -\left\vert y\right\vert \leq y\leq \left\vert y\right\vert }

after adding,

{\displaystyle -(\left\vert x\right\vert +\left\vert y\right\vert )\leq x+y\leq \left\vert x\right\vert +\left\vert y\right\vert }

Use the fact that ( with b replaced by x + y and a ), you have

{\displaystyle \left\vert b\right\vert \leq a\Leftrightarrow -a\leq b\leq a}{\displaystyle \left\vert x\right\vert +\left\vert y\right\vert }
{\displaystyle |x+y|\leq |x|+|y|}

In terms of the size of individual numbers, the triangle inequality is useful in mathematical analysis to determine the best upper estimate on the size of the sum of two numbers.

There is also a reduced approximation, which can be found using the reverse triangle inequality which states that for any real numbers x and y :

|x-y|\geq {\bigg |}|x|-|y|{\bigg |}.

Interior Products Ideal as in an interior product space. If the criterion arises from an inner product (as in the case of Euclidean spaces), then the triangle inequality follows from the Cauchy–Schwarz inequality: given vectors and , and representing the inner product as :

{\displaystyle \|x+y\|^{2}}{\displaystyle =\langle x+y,x+y\rangle }
	=\|x\|^{2}+\angle x,y\rangle +\angle y,x\rangle +\|y\|^{2}
	\leq \|x\|^{2}+2|\langle x,y\rangle |+\|y\|^{2}
	\leq \|x\|^{2}+2\|x\|\|y\|+\|y\|^{2}(by ~~the~~ Cauchy~~–Schwarz~~ inequality)

The Cauchy–Schwarz inequality turns into an equality if and only if x and y are linearly dependent. The inequality turns into an equality for linearly dependent if and only if one of the vectors x or y is a non- negative scalar of the other .

\langle x,y\rangle +\langle y,x\rangle \leq 2|\langle x,y\rangle |xy

Taking the square root of the final result gives the triangle inequality.

  • P-criteria: The commonly used criterion is the p- criteria:
\|x\|_{p}=\left(\sum _{i=1}^{n}|x_{i}|^{p}\right)^{1/p}\ ,

where i are the components of the vector x . For p = 2 the p -norm becomes the Euclidean ideal :

\|x\|_{2}=\left(\sum _{i=1}^{n}|x_{i}|^{2}\right)^{1/2}=\left(\sum _{i=1}^{n}x_{i}^{2}\right)^{1/2}\ ,

Which is Pythagoras’ theorem in n -dimensions, a very special case analogous to an inner product criterion. Except in the case p = 2 , the p -value is not an inner product criterion , as it does not satisfy the parallelogram law. The triangle inequality for normal values ​​of p is called Minkowski’s inequality. [18] This takes the form:

\|x+y\|_{p}\leq \|x\|_{p}+\|y\|_{p}\ .

Metric space

In a metric space M with metric d , the triangle inequality requires a distance:

d(x,\ z)\leq d(x,\ y)+d(y,\ z)\ ,

For all x , y , z in M ​​. That is,the distance from x to z is at most equal to the sum of the distance from x to y and the distance from y to z .

The triangle inequality accounts for most of the interesting structure on a metric space, namely convergence. This is because the remaining requirements for the metric are relatively simple in comparison. For example, the fact that any transformation sequence in a metric space is a Cauchy sequence is a direct consequence of the triangle inequality, because if we choose any n and am such that d ( n , x ) < / 2 and d ( m , x ) < / 2 , where > 0If the triangle inequality is given and used (as the definition of a limit in a metric space) is arbitrary, by d ( n , m ) d ( n , x ) + d ( m ) , x ) < / 2 + / 2 = , so that the sequence { x } is , by definition, a Cauchy sequence. This version of the triangle inequality where a metric is induced by means of the normed vector spaces in the case stated above reduces to a d ( x , y ) x – y , along the x – y point The vector is pointing to y for x .

Opposite triangle inequality

The reverse triangle inequality is a primary consequence of the triangle inequality that gives the lower bound rather than the upper bound. For plane geometry, the statement is: [19]Any side of a triangle is greater than the difference of the other two sides .

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

{\bigg |}\|x\|-\|y\|{\bigg |}\leq \|x-y\|,

or for metric spaces, | d ( y , x ) – d ( x , z )| d ( y , z ) . This implies that the normals are the distance function as well as the Lipschitz constant with the Lipschitz constant of 1 , and are therefore in particular uniformly continuous.

\|\cdot \|{\displaystyle d(x,\cdot )}

The proof for the inverted triangle uses the regular triangle inequality, and :

{\displaystyle \|yx\|=\|{-}1(xy)\|=|{-}1|\cdot \|xy\|=\|xy\|}
\|x\|=\|(x-y)+y\|\leq \|x-y\|+\|y\|\Rightarrow \|x\|-\|y\|\leq \|x-y\|,
\|y\|=\|(y-x)+x\|\leq \|y-x\|+\|x\|\Rightarrow \|x\|-\|y\|\geq -\|x-y\|,

Combining these two statements, we get:

-\|x-y\|\leq \|x\|-\|y\|\leq \|x-y\|\Rightarrow {\bigg |}\|x\|-\|y\|{\bigg |}\leq \|x-y\|.

Minkowski reversal in space

The Minkowski space metric is not positive-definite, which means that either the sign can be or be missing, even if the vector x is non-zero. Furthermore, if x and y are both equal vectors of time that lie in the future light cone, then the triangle inequality is reversed:

 {\displaystyle \|x\|^{2}=\eta _{\mu \nu }x^{\mu }x^{\nu }}
{\displaystyle \ | x + y \ | \geq \ | x \ | + \ | y \ |.}

A physical example of this inequality is the twin paradox in special relativity. The same inverse form of the inequality holds if both vectors lie in the previous light cone, and if one or both are zero vectors. The result holds for any n 1 in n + 1 dimensions. If the plane defined by x and y is isotope (and therefore a Euclidean subspace), then the general triangle inequality occurs.

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