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where the length z of the third side has been replaced by the vector sum x + y. When x and y are real numbers, they can be viewed as vectors in R1, and the triangle inequality expresses a relationship between absolute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.
See the attached image for my attempt. My main concern is can I assume that y > x prove it for that case and then show it is equal if y = x.
My whole proof is centered around y > x so if i cannot make that assumption then I have to start over. Let me know your thoughts. Thanks in advance for...
I have just read about a mathematical conjecture that has presumably been debunked, the bunkbed conjecture. The Wikipedia link hasn't been updated since I wrote this post. The preprint reads
There is a YouTube video (##\approx 15## min.) that explains the situation quite well.
Besides...
Been stuck on this one for a while now.
Base case is easy, n=1, we have x <=1/3, so trivially 1-x>= 2/3 and we are done.
The issue is with the inductive step, I don't know how to use the hint, infact I am struggling to understand what is meant by the hint.
Any help (or a full solution) would...
Hi,
I have problem to prove that the following inequality holds
I thought of the following, since it is a convex function and ##x_1 < x_2 <x_3## applies, I started from the following inequality ##f(x_2) \leq f(x_3)## and transformed it further
$$f(x_2) \leq f(x_3)$$
$$f(x_2)-f(x_1) \leq...
Just went through this...steps pretty clear. I refreshed on Riemann integrals { sum of rectangles approximate area under curves}. My question is on the highlighted part in Red. The approximation of area under curve may be smaller or larger than the actual value. Thus the inequality may be ##<##...
I am interested in an algebraic approach.
My lines are as follows;
##\dfrac{(x+1)(x+4)}{(x-1)(x-2)} -2<0##
##\dfrac{(x^2+5x+4) - 2(x-1)(x-2)}{(x-1)(x-2)} <0##
The denominator will give us the vertical asymptotes ##x=1## and ##x=2##
The numerator gives us,
##x^2+5x+4-2x^2+6x-4 <0##...
Q. 21
Allow me to post using phone... Will amend/ show working when i get hold of computer.In my working I have
##5x^2 -cx^2-4x-2<0##
##56-8c<0##
...
##c>7##
Textbook says ##7>c## ...who's fooling who here? 😊
I have seen my mistake. I am wrong...put in wrong inequality sign...
Should be...
The whole solution is a bit long, which I'll attach but the part I'm stuck at is, assuming everything else above it is correct, is
4 < (log x - 3)(8-log x)
Note ; inequalities aren't technically taught yet in the course, so please try to make the solution not go too deep into that. If that...
Book answer is ##\qquad a≥0\qquad x={ 7\over 9}\;a\ \ \lor\ \ x={ 13\over 14}\;a\ \ ##but I fail to see how to get there !
Stunned by an 1886 dutch high school exam exercise. Hats off for the 17 year olds that did it !
##\ ##
Hi there, i must preface with saying my understanding of the problem is limited to undergraduate quantum mechanics, since my spacialization is chemistry. I know the basic principles, like hilbert spaces, vector bases etc... I'm only asking this question because i genuinely want to understand...
Was trying to understand the inequality test. The only article ever that I've found that explains it simply is the 1981 article, Bringing home the atomic world: Quantum Mysteries For Anybody. All other explanations require trust and understanding of polarisation, which is a huge deal.
So i now...
Hi
If i have the inequality , 9 > x2 then i know the answer is , -3 < x < +3 but my confusion lies in the following ; if i take the square root of both sides of the inequality i get , ± 3 > ±x
Is that correct ? If so , it leads to the following solutions
x < 3 , x > -3 , x < -3 , x >3 ...
First lets focus on ##|x|## which is defined as distance between ##x##and ##0##. But if we look into it closely
$$13=|-11-2|$$ which is distance between -11 and 2 but $$13=|11-(-2)|$$ which means this is distance between 11 and -2. Which is it?
In the same way $$x=|x-0|$$ is distance between 0...
For this,
I don't understand how they got from (1) to (2)? Dose someone please know what binary operation allows for that?
I also don't understand how they algebraically got from line (2) to (3).
Many thanks!
Let ##\Omega## here be ##\Omega=\sqrt{-u}##, in which it is not difficult to realize that ##\Omega ## is real if ##u<0##; imaginary, if ##u>0##. Now, suppose further that ##u=(a-b)^2## with ##a<0## and ##b>0## real numbers. Bearing this in mind, I want to demonstrate that ##\Omega## is real. To...
Lemma
Let f be a bounded function on [a,b]. If P & Q are partitions of [a,b] and P ##\subseteq## Q , then
L(f,P) ##\leq## L(f,Q) ##\leq## U(f,Q) ##\leq## U(f,P) .
Question is "How can P have bigger upper darboux sum than Q while it is a subset of Q"
Im a having trouble understanding how this exactly works.
$$ |x^2 - 4| < |x^2+2| $$ So I know the usual thing to do when you have absolute values,here it is even simpler since the right part of the inequality is always positive so I just have these 2 cases.
1. ## x^2-4 >= 0 ## and 2. ## x^2-4...
Hi,
Recently I studied triangle inequality and the proof using textbook precalculus by David Cohen.
My question is whats the benefit of this inequality ? One benefit I found is to solve inequality of the form |x+a| + |x+b| < c which make the solution much easier than taking cases. I assume this...
The Bell inequality requires three conditions, A, B and C that can have two values (pass or fail, say). In the Aspect experiment A defines a plane, B a plane of A rotated by 22.5 degrees, while C is A rotated by 45 degrees. We take joint probabilities, and two are A+.B-, and B+.C-, and from the...
Let ##F## be a field. If ##A \in M_{n\times k}(F)## and ##B\in M_{k\times n}(F)##, show that $$\operatorname{rank}(A) + \operatorname{rank}(B) - k \le \operatorname{rank}(AB) \le \min\{\operatorname{rank}(A), \operatorname{rank}(B)\}$$
The assignment says proof by induction is possible, I cannot figure out how this is supposed to work out. Does somebody know the name of this by any chance? Seeing a derivation might help come up with an idea for a proof. Thank you everybody.
I would like to show that a LLL-reduced basis satisfies the following property (Reference):
My Idea:
I also have a first approach for the part ##dist(H,b_i) \leq || b_i ||## of the inequality, which I want to present here based on a picture, which is used to explain my thought:
So based...
Problem Statement : Solve the inequality : ##\left( \dfrac{1}{3} \right)^x<9##.
Attempts: I copy and paste my attempt below using Autodesk Sketchbook##^{\circledR}##. The two attempts are shown in colours black and blue.
Issue : On checking, the first attempt in black turns out to be...
Problem Statement : I copy and paste the problem as it appeared in the text to the right.
Attempt (mine) : I copy and paste my attempt using Autodesk Sketchbook##^{\circledR}## below. I hope the writing is legible.
My answer : I have three answers and confused as to which of them hold...
I just read an article in Quantum Magazine about "unitary" results and how this is tied to looking at the reversibility of quantum events.
It provided an easy-to-understand mechanism for tracking the effects of adding information to a fictional universe. The example they gave for detecting a...
Here is my attempt (Note:
## \left| \int_{C} f \left( z \right) \, dz \right| \leq \left| \int_C udx -vdy +ivdx +iudy \right|##
##= \left| \int_{C} \left( u+iv, -v +iu \right) \cdot \left(dx, dy \right) \right| ##
Here I am going to surround the above expression with another set of...
Problem statement : Let me copy and paste the problem as it appears in the text.
Attempt : From the "Relevant Equations" given above, we can compare to see that ##a-1 = -1## and ##a^2+2=3##. These lead (after some algebra) to the three values of ##\boxed{a=0, \pm 1}##.
Issue : The book has a...
My first attempt was ##... + n^{2} + (n+1)^{2} > \frac {1}{3} n^{3} + (n+1)^{2}##
then we must show that ##\frac {1}{3} n^{3} + (n+1)^{2} > \frac {1}{3} (n+1)^{3}##
We evaluate both sides and see that the LHS is indeed bigger than RHS. However, this solution is inconsistent so I am asking for...
Summary: Can these two equations be solved for x like a system of linear inequalities, and how?
##x- 2y \le 54##
##x + y \ge 93##
We start with
##x- 2y \le 54##
##x + y \ge 93##
Multiplying the second equation by 2, we have ##2x + 2y \ge 184##. We cannot seem to cancel the y out with the...
Hi. I was reading Lighthill, Introduction to Fourier Analysis and Generalised Functions and in page 17 there is an example/proof where I can't make sense of the following step:
$$
\left| \int_{-\infty}^{+\infty} f_n(x)(g(x)-g(0)) \, \mathrm{d}x \right| \le
\max{ \left| g'(x) \right| }...
Problem Statement : I copy and paste the problem from the text to the right.
Attempt (mine) : Given the inequality ##\dfrac{x}{x+2}\le \dfrac{1}{|x|}##. We see immediately that ##x\ne 0, -2##. At the same time, since ##|x|\ge 0\Rightarrow \frac{x}{x+2}\ge 0##.
Now if ##\frac{x}{x+2}\le...
Problem statement : Let me copy and paste the problem as it appears in the text on the right.Attempt (myself) : By looking at ##\large{\sqrt{x+2}\ge x}##, from my Relevant Equations above, we have the following :
1. Outcome ##\mathbf{x \ge 0}##, since square roots are always positive.
2...
Find solution here;
Ok i just want clarity for part (a),
My approach is as follows, since we want positive integer values that satisfy the problem then,
##\dfrac {n^2-1}{2}≥1## I had earlier thought of ##\dfrac {n^2-1}{2}≥0## but realized that ##0## is an integer yes but its not a positive...
I have a two-variable inequality and wish to make a graph of the regions in which it is satisfied. Is any such took available online for free?
A great many free online graphing calculators are available, but I expect the great majority won't do what I want.
Specifically I want to find the...
I don't need an answer (although I don't have sadly, it's from a test).
I need just a tip on how to start it...
i cannot use Taylor in here (##\ln(x)## is not Taylor function), therefore, its only MVT, but I don't know which point I should try... since I must get the annoying ##\ln(x)##...
I'm trying to prove the following:
##wt(x+y) \leq wt(x) + wt(y)##, where "wt(x)" is referring to the weight of a specific code word.
Proof:
For two code words ##x, y \in F^{n}_2##, we have the inequalities ##0 \leq wt(x)## and ##0 \leq wt(y)##. Adding these together, we have ##0 \leq wt(x) +...
Trying to get the last one, but not able to do it. What am I missing. I know that I can't but [-2,infinity] or [3,infinity] because for -2, and 3 y=0 and is not > 0. Any help appreciated.
The statements holds for the case ##n=1##
\begin{align} 1+x \leq & 1+x+\frac{1}{2}\cdot 0\cdot x^2\\
=&1+x \end{align}
Assume the statement holds true for ##n=k##
$$(1+x)^k\geq 1+kx+\frac{k(k-1)}{2} x^2$$
Then, for ##n=k+1##, we have the following
\begin{align} (1+x)^k\cdot (1+x)\geq&...
This is the question * consider the highlighted question only *with its solution shown (from textbook);
My approach is as follows (alternative method),
##|\frac {5}{2x-3}| ##< ## 1##
Let, ##\frac {5}{2x-3} ##⋅##\frac {5}{2x-3}##=##1##
→##x^2-3x-4=0##
##(x+1)(x-4)=0##
it...
Just to remind, ##C_\ell## is the variance of random variables ##a_{\ell m}## following a Gaussian PDF (in spherical harmonics of Legendre) :
##C_{\ell}=\left\langle a_{l m}^{2}\right\rangle=\frac{1}{2 \ell+1} \sum_{m=-\ell}^{\ell} a_{\ell m}^{2}=\operatorname{Var}\left(a_{l m}\right)##
1)...
Let $x, y, z$ be length of the side of a triangle such that $\sqrt{x} + \sqrt{y} + \sqrt{z} = 1.$
Prove $|x^{2} + y^{2} + z^{2} - 2\left( xy+yz+xz\right)| \le \frac{1}{27}$.
What is the meaning of this proof? What is the meaning of last statement of this proof? How to prove lemma (7.1)? or How to answer problem 1 given below?
I've no political agenda. This is strictly a math question about optimization. Those addressing it should do so strictly from a mathematical perspective. Thank you!
If the world simply had one global currency, like Bitcoin, and everyone had an account, then the sum total of all accounts...