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):
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x
+
y
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≤
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x
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+
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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where the length z of the third side has been replaced by the vector sum x + y. When x and y are real numbers, they can be viewed as vectors in R1, and the triangle inequality expresses a relationship between absolute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.
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...
Let $a,\,b$ and $c$ be the side lengths of a triangle. Prove that $\dfrac{a}{\sqrt[3]{4b^3+4c^3}}+\dfrac{c}{\sqrt[3]{4a^3+4b^3}}+\dfrac{a}{\sqrt[3]{4b^3+4c^3}}<2$.
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)...
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...
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...
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...
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}
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...
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}| =...
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 ${y}_{n}\to {x}^{*}$
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
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...
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...
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 ?
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?
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...
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.
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...
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...
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...
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...
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)...
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...
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.
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...
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.
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| ≤...
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...
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...
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-...
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...
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...
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.
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...
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...
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| =...
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...
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)...
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...
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) <=...
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...
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...
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...