What is Triangle inequality: Definition and 102 Discussions

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. This statement permits the inclusion of degenerate triangles, but some authors, especially those writing about elementary geometry, will exclude this possibility, thus leaving out the possibility of equality. If x, y, and z are the lengths of the sides of the triangle, with no side being greater than z, then the triangle inequality states that




z

x
+
y
,


{\displaystyle z\leq x+y,}
with equality only in the degenerate case of a triangle with zero area.
In Euclidean geometry and some other geometries, the triangle inequality is a theorem about distances, and it is written using vectors and vector lengths (norms):






x

+

y





x


+


y


,


{\displaystyle \|\mathbf {x} +\mathbf {y} \|\leq \|\mathbf {x} \|+\|\mathbf {y} \|,}
where the length z of the third side has been replaced by the vector sum x + y. When x and y are real numbers, they can be viewed as vectors in R1, and the triangle inequality expresses a relationship between absolute values.
In Euclidean geometry, for right triangles the triangle inequality is a consequence of the Pythagorean theorem, and for general triangles, a consequence of the law of cosines, although it may be proven without these theorems. The inequality can be viewed intuitively in either R2 or R3. The figure at the right shows three examples beginning with clear inequality (top) and approaching equality (bottom). In the Euclidean case, equality occurs only if the triangle has a 180° angle and two 0° angles, making the three vertices collinear, as shown in the bottom example. 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 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 [0, π]) with those endpoints.The triangle inequality is a defining property of norms and measures of distance. This 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, Euclidean spaces, the Lp spaces (p ≥ 1), and inner product spaces.

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  1. M

    I Usage of triangle inequality?

    Hi, Recently I studied triangle inequality and the proof using textbook precalculus by David Cohen. My question is whats the benefit of this inequality ? One benefit I found is to solve inequality of the form |x+a| + |x+b| < c which make the solution much easier than taking cases. I assume this...
  2. S

    MHB Proof of Triangle Inequality for $n$ Natural Numbers

    Prove for all $n\in N$ $\dfrac{|a_1+...a_n|}{1+|a_1+...+a_n|}\leq\dfrac{|a_1|}{1+|a_1|}+...\dfrac{|a_n|}{1+|a_n|}$
  3. anemone

    MHB Prove Triangle Inequality: $\frac{a}{\sqrt[3]{4b^3+4c^3}}+...<2$

    Let $a,\,b$ and $c$ be the side lengths of a triangle. Prove that $\dfrac{a}{\sqrt[3]{4b^3+4c^3}}+\dfrac{c}{\sqrt[3]{4a^3+4b^3}}+\dfrac{a}{\sqrt[3]{4b^3+4c^3}}<2$.
  4. binbagsss

    Triangle Inequality: use to prove convergence

    Homework Statement Attached I understand the first bound but not the second. I am fine with the rest of the derivation that follows after these bounds, Homework Equations I have this as the triangle inequality with a '+' sign enabling me to bound from above: ##|x+y| \leq |x|+|y| ## (1)...
  5. L

    Inner Product, Triangle and Cauchy Schwarz Inequalities

    Homework Statement Homework Equations I am not sure. I have not seen the triangle inequality for inner products, nor the Cauchy-Schwarz Inequality for the inner product. The only thing that my lecture notes and textbook show is the axioms for general inner products, the definition of norm...
  6. Mr Davis 97

    Proving the Triangle Inequality: ##|a-b| < \epsilon##

    Homework Statement If ##\forall \epsilon > 0 ## it follows that ##|a-b| < \epsilon##, then ##a=b##. Homework EquationsThe Attempt at a Solution Proof by contraposition. Suppose that ##a \neq b##. We need to show that ##\exists \epsilon > 0## such that ##|a-b| \ge \epsilon##. Well, let...
  7. lfdahl

    MHB Prove Triangle Inequality: $\sum_{cyc} \sin A$

    Prove, that for any triangle:\[ \sum_{cyc}\sin A - \prod_{cyc}\sin A \ge \sum_{cyc}\sin^3 A \]
  8. T

    Triangle inequalities

    Just wondering if anyone could confirm if I've headed in the right direction with these (a) Prove the triangular inequality: |x + y| ≤ |x| + |y|. (b) Use triangular inequality to prove |x − y| ≥ ||x| − |y||. (c) Show that if |x − a| < c/2 and |y − b| < c/2 then |(x + y) − (a + b)| < c. So for...
  9. anemone

    MHB Right-Angled Triangle Inequality

    Show that if $a,\,b$ and $c$ are the lengths of the sides of a right triangle with hypotenuse $c$, then \frac{(c − a)(c − b)}{(c + a)(c + b)}\le 17 − 12\sqrt{2}
  10. L

    A Triangle inequality question

    In the derivation of triangle inequality |(x,y)| \leq ||x|| ||y|| one use some ##z=x-ty## where ##t## is real number. And then from ##(z,z) \geq 0## one gets quadratic inequality ||x||^2+||y||^2t^2-2tRe(x,y) \geq 0 And from here they said that discriminant of quadratic equation D=4(Re(x,y))^2-4...
  11. B

    Triangle inequality implies nonnegative scalar multiple

    I'm not really sure if this is true, which is why I want your opinion. I have been trying to prove it, but it will help me a lot if someone can confirm this. Let ## v_{1}, v_{2} ... v_{n} ## be vectors in a complex inner product space ##V##. Suppose that ## | v_{1} + v_{2} +...+ v_{n}| =...
  12. O

    MHB Triangle Inequality and Convergence of ${y}_{n}$

    Let ${y}_{n}$ be a arbitrary sequence in X metric space and ${y}_{m+1}$ convergent to ${x}^{*}$ in X...İn this case by using triangle inequality can we say that ${y}_{n}\to {x}^{*}$
  13. lucasLima

    Help proving triangle inequality for metric spaces

    So, i need to proof the triangle inequality ( d(x,y)<=d(x,z)+d(z,y) ) for the distance below But I'm stuck at In those fractions i need Xk-Zk and Zk-Yk in the denominators, not Xk-Yk and Xk-Yk. Thanks in advance
  14. O

    MHB Generalized triangle inequality in b-metric spaces

    How is the generalized triangle inequality in b-metric spaces ? I find something...But I wonder your opinion...Thank you for your attention... Especially if you write for n,m>0 m>n $d({x}_{n},{x}_{m})$$\le$..... I will be happy...
  15. O

    MHB Triangle inequality in b-metric spaces

    Let $X$ be a non-empty set and let $s\ge1$ be a given real number. A function $d:$ X $\times$ X$\to$ ${R}^{+}$ , is called a b-metric provided that, for all x,y,z $\in$ X, 1) d(x,y)=0 iff x=y, 2)d(x,y)=d(y,x), 3)d(x,z)$\le$s[d(x,y)+d(y,z)]. A pair (X,d) is called b-metric space. İt is clear...
  16. ognik

    MHB Reverse triangle inequality with a + sign

    Thought I knew this, but am confused by the following example: Show $ |z^3 - 5iz + 4| \ge 8 $ The example goes on: $ |z^3 - 5iz + 4| \ge ||z^3 - 5iz| - |4|| $, using the reverse triangle inequality It's probably right, but I don't get why the +4 can just be made into a -4 ?
  17. J

    The triangle inequality in CHSH, where is the triangle?

    http://en.wikipedia.org/wiki/CHSH_inequality#Bell.27s_1971_derivation The last step of the CHSH inequality derivation is to apply the triangle inequality. I see there are relative polarization angles, but I don't see any sides have defined length to make up a triangle. Where is the triangle?
  18. C

    Triangle inequality proof in Spivak's calculus

    So hi, there's one little thing which I'm not understanding in the proof. After the inequality Spivak considers the two expressions to be equal. Why?!? I just don't see why we can't continue with the inequality and when we have factorized the identity to (|a|+|b|)^2 we can just replace...
  19. Albert1

    MHB Prove a triangle inequality

    Triangle ABC with side lengths a,b,c please prove : $ \sqrt {ab}+\sqrt {bc}+\sqrt {ca}\leq a+b+c<2\sqrt {ab}+2\sqrt {bc}+2\sqrt {ca}$
  20. T

    Triangle Inequality in 'Linear Algebra Done Right'

    I'm stuck on one aspect of the proof on page 105 of the 2nd edition. Equation 6.13 is necessary for the inequality to be an equality as it says but they never seem to account for inequality 6.11. Specifically, I don't see how this satisfies 2 Re<u,v> = 2 |<u,v>| Thanks for any guidance.
  21. B

    Variation of the triangle inequality on arbitrary normed spaces

    The following inequality can easily be proved on ##ℝ## : ## ||x|-|y|| \leq |x-y| ## I was wondering if it extends to arbitrary normed linear spaces, since I can't seem to prove it using the axioms for linear spaces. (I can however, prove it using the definition of the norm on ##ℝ## by using...
  22. B

    Triangle Inequality Proof

    Hello all, I am currently reading about the triangle inequality, from this article http://people.sju.edu/~pklingsb/cs.triang.pdf I am curious, how does the equality transform into an inequality? Does it take on this change because one takes the absolute value of 2uv? Because before the...
  23. Seydlitz

    Triangle Inequality Question

    I'm beginning to read Spivak's Calculus 3ed, and everything is smooth until I reach page 12. My question is marked, between line 2 and 3. Why there's such sign change suddenly? In fact I tried with simple line 4 case and it's not in fact equal. I'm assuming that a and b is valid for all...
  24. B

    Triangle Inequality Proof

    Homework Statement Use the triangle inequality to prove that \left| s_n - s \right| < 1 \implies \left| s_n \right| < \left| s \right| +1 Homework Equations The triangle inequality states that \left| a-b \right| \leq \left| a-c \right| + \left| c-b \right| The Attempt at a Solution...
  25. Albert1

    MHB Proof of Triangle Inequality: a+b-c, b+c-a, c+a-b

    Let a, b, c be the lengths of the sides of a triangle. Prove that: $\sqrt{a+b-c}$+$\sqrt{b+c-a}$+$\sqrt{c+a-b}\leq\sqrt{a}+\sqrt{b}+\sqrt{c}$
  26. S

    Using the generalized triangle inequality

    Homework Statement Using the generalized triangle inequality, prove |d(x,y) - d(z,w)| ≤ d(x,z) + d(y,w) Homework Equations d(x,y) is a metric triangle inequality: d(x,y) ≤ d(x,z) + d(z,y) The Attempt at a Solution I know that this needs to be proved with cases: a) d(x,y) - d(z,w)...
  27. S

    Triangle Inequality Proving: Use Sine Law & Find Solution

    Homework Statement Prove the following inequality for any triangle that has sides a, b, and c. -1<\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-\frac{a}{c}-\frac{c}{b}<1 Homework Equations The Attempt at a Solution I think we have to use sine or cosine at a certain point because...
  28. 8

    Scalar product to prove triangle inequality?

    Homework Statement From the inequality |a.b| <= |a||b| prove the triangle inequality: |a+b| <= |a| + |b| Homework Equations a.b = |a|b| cos theta The Attempt at a Solution Making a triangle where side c = a+b. Don't know how to approach the question. Thanks.
  29. S

    Triangle Inequality Proof

    Homework Statement Prove llxl-lyll≤lx-yl (The triangle inequality: la+bl≤lal+lbl) The Attempt at a Solution For the first part, I assumed lxl≥lyl: lxl=l(x-y)+yl Then, by Triangle Inequality l(x+y)+yl≤l(x-y)l+lyl So, lxl≤l(x-y)l+lyl Subtract lyl from both sides to...
  30. B

    Triangle inequality for a normalized absolute distance

    Hi, can you please give me some hints to show that \frac{|a-b|}{1+|a|+|b|} \leq \frac{|a-c|}{1+|a|+|c|}+\frac{|c-b|}{1+|c|+|b|}, \forall a, b, c \in \mathbb{R}. I tried to get this from |a-b| \leq |a-c|+|c-b|, \forall a, b, c \in \mathbb{R}, but I couldn't succeed. Thank you.
  31. D

    Proof of the triangle inequality

    I am familiar with the proof for the following variant of the triangle inequality: |x+y| ≤ |x|+|y| However, I do not understand the process of proving that there is an equivalent inequality for an arbitrary number of terms, in the following fashion: |x_1+x_2+...+x_n| ≤...
  32. Ryuzaki

    Triangle Inequality and the Triangle Law of Vector Addition

    The triangle inequality states that, the sum of any two sides of a triangle must be greater than the third side of the triangle. But the triangle law of vector addition states that if we can represent two vectors as the two sides of a triangle in one order ,the third side of the triangle...
  33. C

    Double slit experiment violates triangle inequality?

    Imagine a light source, double-slit, and a curved screen in vacuum, shaped so that all parts of the interference pattern are created simultaneously. Define distance as proportional to the time light requires to reach a point. Detectors at each slit can be operating or not. Call the source S...
  34. C

    Generalized triangle inequality

    Homework Statement Show that |x_1 + x_2 + · · · + x_n | ≤ |x_1 | + |x_2 | + · · · + |x_n | for any numbers x_1 , x_2 , . . . , x_n Homework Equations |x_1 + x_2| ≤ |x_1| + |x_2| (Triangle inequality)The Attempt at a Solution I tried using the principle of induction here, but to no avail...
  35. J

    Baby Rudin Proof of Theorem 1.33 (e) - Triangle Inequality

    Hi everyone, I have a question on Rudin's proof of Theorem 1.33 part e. Here he prove the following statement: The absolute value of z+w is equal or smaller than the absolute value of z plus the absolute value of w -Yes, is the triangle inequality, where z and w are both complex numbers-...
  36. G

    CHSH and the triangle inequality

    Hello everybody, I've been trying to understand the CHSH proof as it is listed on Wikipedia: http://en.wikipedia.org/wiki/CHSH_inequality I got to this without any problem: E(a, b) - E(a, b^\prime) = \int \underline {A}(a, \lambda)\underline {B}(b, \lambda)[1 \pm \underline {A}(a^\prime...
  37. J

    Proving the triangle inequality property of the distance between sets

    Proving the "triangle inequality" property of the distance between sets Here's the problem and how far I've gotten on it: If you are unfamiliar with that notation, S(A, B) = (A \ B) U (B \ A), which is the symmetric difference. And D(A, B) = m^*(S(A, B)), which is the outer measure of...
  38. D

    MHB Triangle inequality question

    I am trying to show $|(n+z)^2|\leq (n -|z|)^2$ where is complex $|(n+z)^2| = |n^2 + 2nz + z^2| \leq n^2 + 2n|z| + |z|^2$ But I can't figure out the connection for the final piece.
  39. Advent

    Triangle inequality for complex numbers: sketch of proof

    Homework Statement Show that if z_1,z_2 \in \mathbb{C} then |z_1+z_2| \leq |z_1| + |z_2| Homework Equations Above. The Attempt at a Solution I tried by explicit calculation, with obvious notation for a,b and c: my frist claim is not that the triangle inequality holds, just that...
  40. T

    Triangle Inequality for a Metric

    Homework Statement Prove the triangle inequality for the following metric d d\big((x_1, x_2), (y_1, y_2)\big) = \begin{cases} |x_2| + |y_2| + |x_1 - y_1| & \text{if } x_1 \neq y_1 \\ |x_2 - y_2| & \text{if } x_1 = y_1 \end{cases}, where x_1, x_2, y_1, y_2 \in \mathbb{R}...
  41. T

    Using Triangle Inequality to find a magnitude

    Homework Statement Let a, x, and y be real numbers and let E > 0. Suppose that |x-a|< E and |y-a|< E. Use the Triangle Inequality to find an estimate for the magnitude |x-y|. Homework Equations The Triangle Inequality states that |a+b| <= |a| + |b| is valid for all real numbers a and...
  42. H

    Triangle Inequality for integrals proof

    Homework Statement What I want to show is this: ∫|x+y| ≤ ∫|x| + ∫|y| Homework Equations |x+y| ≤ |x| + |y| The Attempt at a Solution So I thought if I used the triangle inequality I could get to something along the lines of: Lets g belong to the real numbers ∫|x+y| =...
  43. L

    Real Analysis problem (easy), Triangle inequality

    Homework Statement > a[1], a[2], a[3], .. , a[n] are arbitrary real numbers, prove that; abs(sum(a[i], i = 1 .. n)) <= sum(abs(a[i]), i = 1 .. n) Homework Equations The Attempt at a Solution I have uploaded my attempt as a pdf file, since I'm not too familiar with the...
  44. C

    Prove using the Triangle Inequality

    Homework Statement Show that: (|x+y|)/(1+|x+y|) ≤ ((|x|)/(1+|x|)) + ((|y|)/(1+|y|)) Homework Equations You are given the triangle inequality: |x+y| ≤ |x| + |y| The Attempt at a Solution (This is done from the result, as I haven't been able to find the starting point)...
  45. R

    Epsilon-Delta Proof for Continuity of f + 2g at x = a

    Homework Statement Part of an \epsilon-\delta proof about whether or not f + 2g is continuous at x = a provided that f and g are continuous at x = a The Attempt at a Solution I've got the proof (I hope), but I'm uncertain about whether I can do the following...
  46. P

    Triangle Inequality Proof

    Im curios as to why the inquality is ||x+y||\leq||X||+||y|| but the end of the proof is =(||x||+||Y||)^2 where does the less than symbol disappear too
  47. M

    Reverse Triangle Inequality

    Homework Statement I'm reading the proof for the reverse triangle inequality, but I don't understand what is meant by "by symmetry" Homework Equations The Attempt at a Solution (X,d) is a metric space prove: |d(x,y) - d(x,z)| <= d(z,y) The triangle inequality d(x,y) <=...
  48. S

    Proving the Triangle Inequality Theorem using Coordinates

    Homework Statement Prove the Triangle Inequality Theorum using the coordinate system. Homework Equations The corners of the triangles will be at (x1,y1), (x2, y2), (x3,y3) The Attempt at a Solution The proof that I know is proving that |x+y|<=|x|+|y|: -|x|<x<|x|, and...
  49. C

    Metric Spaces, Triangle Inequality

    I have the following question on metric spaces Let (X,d) be a metric space and x1,x2,...,xn ∈ X. Show that d(x1, xn) ≤ d(x1, x2) + · · · + d(xn−1, xn2 ), and d(x1, x3) ≥ |d(x1, x2) − d(x2, x3)|. So the first part is simply a statement of the triangle inequality. However, the metric...
  50. G

    Elementary Analysis, Triangle Inequality Help

    Homework Statement Prove that ||a|-|b||\leq |a-b| for all a,b in the reals Homework Equations I know we have to use the triangle inequality, which states: |a+b|\leq |a|+|b|. Also, we proved in another problem that |b|\leq a iff -a\leqb\leqa The Attempt at a Solution Using the...
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