reverse triangle inequality proof

Proof. Move $|x|$ to the right hand side in the first inequality and $|y|$ to the right hand side in the second inequality. In a normed vector space V, one of the defining properties of the norm is the triangle inequality: The Triangle Inequality for Inner Product Spaces. If you think about $x$ and $y$ as points in $\mathbb{C}$, on the left side you’re keeping the distance of both the vectors from 0, but making them both lie on the positive real axis (by taking the norm) before finding the distance, which will of course be less than if you just find the distance between them as they are (when they might be opposite each other in the complex plane). Also then . Proof. \begin{equation*} The reverse triangle inequality is an elementary consequence of the triangle inequality that gives lower bounds instead of upper bounds. From absolute value properties, we know that $|y-x| = |x-y|,$ and if $t \ge a$ and $t \ge −a$ then $t \ge |a|$. Proof of the first result is: As then . Observe that there are two (positive) quantities on the left of the $\ge$ sign and one of the right. .net – How to disable postback on an asp Button (System.Web.UI.WebControls.Button). |x-y|=x-y,&x\geq{}y\geq0\\ |x-y|=-x+y=-(x-y),&y\geq{}x\geq0\\ Solution: By the Triangle Inequality, |x−y| = |(x−a)+(a−y)|≤|x−a|+|a−y|≤ + =2 Thus |x−y| < 2 . Use the triangle inequality to see that 0 ja bj= ja a n+ a n bj ja a nj+ ja n bj. $$ Hope this helps and please give me feedback, so I can improve my skills. Antinorms and semi-antinorms M. Moszynsk a and W.-D. Richter Abstract. The main tool used in the proofs is the representation for a power of the farthest distance function as |x|-|y|\ge -|x-y|\;.\tag{2} For plane geometry, the statement is: [19] Any side of a triangle is greater than the difference between the other two sides . In other words, any side of a triangle is larger than the subtracts obtained when the remaining two sides of a triangle are subtracted. We get. (d) jaj 1are such that 1 p + 1 q =1, then xy ≤ xp p + yq q. \begin{align} Geometrically, the triangular inequality is an inequality expressing that the sum of the lengths of two sides of a triangle is longer than the length of the other side as shown in the figure below. |-x-y|=x+y\leq-x+y=-(x-y),&y\geq-x\geq0\\ « Find the area of a parallelogram using diagonals. Interchaning $x\leftrightarrow y$ gives By accessing or using this website, you agree to abide by the Terms of Service and Privacy Policy. \end{align}. Therefore, what we need to prove are (both of) the following: |-x+y|=x-y,&-y\geq-x\geq0\\ For all $x,y\in \mathbb{R}$, the triangle inequality gives We don’t, in general, have $x+(x-y)=y$. The proof is as follows. cr(X) )$, and how certain sentences can be augmented into simpler forms. |x+y|\le|x|+|y|. Proof. The proof is below. Before starting the proof, recall that the triangle inequality says that given a;b2 C ja+ bj jaj+ jbj We can turn this into a lower bound, which we will call the reverse triangle inequal-ity (but it’s not standard) (1) ja+ bj jajj bj by noticing that jaj= j(a+ b) bj ja+ bj+ jbj Also jaj= aand jbj= … Also, … These two results mean that i.e. Hölder's inequality is used to prove the Minkowski inequality, which is the triangle inequality in the space L p (μ), and also to establish that L q (μ) is the dual space of L p (μ) for p ∈ [1, ∞). Section 7-1 : Proof of Various Limit Properties. \end{equation} $$ \left||x|-|y|\right| \leq |x-y|. In this section we are going to prove some of the basic properties and facts about limits that we saw in the Limits chapter. |x|-|y|\leq |x-y| \tag{1}. Rewriting $|x|-|y| \leq |x-y|$ and $||x|-y|| \leq |x-y|$. Combining these two facts together, we get the reverse triangle inequality: WLOG, consider $|x|\ge |y|$. The proof was simple — in a sense — because it did not require us to get creative with any intermediate expressions. Problem 8(a). Then ab 0, so jabj= ab. What is the main concepts going on in this proof? ja+ bj jaj+ jbj. Triangle inequality gives an upper bound 2 , whereas reverse triangle inequalities give lower bounds 2 2 p 2 for general quantum states and 2 2 for classical (or commuting) states. If x+y > 0; then (2) jx+ yj= x+ y jxj+ jyj: On the other hand, if x+ y 0, then (3) jx+ yj= (x+ y) = x y jxj+ jyj: This completes the proof. For first and second triangle inequality, Combining these two statements gives: Triangle inequality giv es an upp er bound 2 − , whereas reverse triangle ine qualities give lower bounds 2 − 2 √ 2 for general quantum states and 2 − 2 for classical (or commuting) \right\}\nonumber\\ $$, On the other hand, the known triangle inequality tells us that “the sum of the absolute values is greater than or equal to the absolute value of the sum”: \begin{array}{ll} https://goo.gl/JQ8Nys Reverse Triangle Inequality Proof. \blacksquare We will now look at a very important theorem known as the triangle inequality for inner product spaces. In the case of a norm vector space, the statement is: The proof for the reverse triangle uses the regular triangle inequality, and. By so-called “first triangle inequality.”. The reverse triangle inequality is an elementary consequence of the triangle inequality that gives lower bounds instead of upper bounds. The truly interested reader can find full proofs in Professor Bhatia’s notes (follow the link above) or in [1]. $$. Hence: \end{equation} The Triangle Inequality can be proved similarly. Hölder's inequality was first found by Leonard James Rogers (Rogers (1888)), and discovered independently by Hölder (1889) Compute |x−y. Privacy policy. |y|+|x-y|\ge |x|\tag{1′} How should I pass multiple parameters to an ASP.Net Web API GET? A vector v 2V is called a unit vector if kvk= 1. Strategy. Now combining $(2)$ with $(1)$, gives (adsbygoogle = window.adsbygoogle || []).push({}); real analysis – Reverse Triangle Inequality Proof. =&|x-y|.\nonumber \begin{equation} which when rearranged gives \end{equation} because $|x-y|=|y-x|$. De nition: Unit Vector Let V be a normed vector space. (Otherwise we just interchange the roles of x and y.) |A|+|B|\ge |A+B|\;\tag{3} PROOF By the triangle inequality, kvk= k(v w) + wk kv wk+ kwk; and the desired conclusion follows. The validity of the reverse triangle inequality in a normed space X. is characterized by the finiteness of what we call the best constant cr(X)associ­ ated with X. Namely for :$x,y,z\in\mathbb{R}, \quad x+(-x)=0$; $x+(y+z)=(x+y)+z$; and $x+y=y+x$. Let’s move on to something more demanding. jjajj bjj ja bj. 8. Reverse (or inverse) triangle inequalities: ka+ bk 2 kak 2 k bk 2 ka+ bk 2 kbk 2 k ak 2 878O (Spring 2015) Introduction to linear algebra January 26, 2017 4 / 22 The reverse triangle inequality is an elementary consequence of the triangle inequality that gives lower bounds instead of upper bounds. The difficult case \end{equation}, $\left| |x|-|y| \right|^2 – |x-y|^2 = \left( |x| – |y| \right)^2 – (x-y)^2 = |x|^2 – 2|x| \cdot |y| +y^2 – x^2 + 2xy-y^2 = 2 (xy-|xy|) \le 0 \Rightarrow \left| |x|-|y| \right| \le |x-y|.$. The paper concerns a biunique correspondence between some pos-itively homogeneous functions on Rn and some star-shaped sets with nonempty interior, symmetric with … This gives the desired result Now we are done by using (3) again. ||x|-|y||\le|x-y|. For real numbers, the formal statement of the inequality is: A corollary of this result, also known as the "reverse triangle inequality", is: Proof. \begin{equation} From absolute value properties, we know that | y − x | = | x − y |, and if t ≥ a and t ≥ − a then t ≥ | a |. It is possible to do a di erent case analysis, e.g. $$ $$ (e)(Reverse Triangle Inequality). \end{equation*} \begin{equation} \bigl||x|-|y|\bigr| (a)Without loss of generality, we consider three cases. |-x+y|=-x-y\leq{}x-y,&-y\geq{}x\geq0\\ Proposition 1 Reverse Triangle Inequality Let V be a normed vector space. Reverse triangle inequality. A Proof of the Reverse Triangle Inequality Let's suppose without loss of generality that ||x|| is no smaller than ||y||. \begin{equation*} Then apply $|x| = |(x-y)+y|\leq |x-y|+|y|$. Triangle inequality Lemma (Triangle inequality) Given a;b 2RN, ka+ bk 2 kak 2 + kbk 2: Proof uses Cauchy-Schwarz inequality (do on board) When does this inequality hold with equality? \begin{equation} a\le M,\quad a\ge -M\;. Reverse Triangle Inequality. $$||x|-|y||=||x-y+y|-|y||\le ||x-y|+|y|-|y||=||x-y||=|x-y|.$$, Explicitly, we have \left\{ Given that we are discussing the reals, $\mathbb{R}$, then the axioms of a field apply. For all a2R, jaj 0. Thus, we get (1′) easily from (3), by setting $A=y$, $B=x-y$. $$ For a nondegenerate triangle, the sum of the lengths of any two sides is strictly greater than the third, thus 2p = a +b +c >2c and so on. |y|-|x| \leq |y-x| Reverse Triangle Inequality Proof Please Subscribe here, thank you!!! |-x+y|=-x+y=-(x-y),&-x\geq-y\geq0\\ (c)(Nonnegativity). |x|+|y-x|\ge |y|\tag{2”} But wait, (2′) is equivalent to (b)(Triangle Inequality). Sas in 7. d(f;g) = max a x b jf(x) g(x)j: This is the continuous equivalent of the sup metric. Suppose |x−a| <, |y −a| <. | x − y | ≥ | x | − | y |. \end{equation*} Let y ≥ 0be fixed and consider the function |-x-y|=-x-y\leq-x+y=-(x-y),&-x\geq{}y\geq0\\ -\left(|x|-|y|\right)\leq |x-y|. Theorem The area of a triangle with given perimeter 2p = a+b+c is maximum if the sides a, b, c are equal. . |-x+y|=x+y\leq{}x-y,&x\geq-y\geq0 The inequality $|a|\le M$ is equivalent to $-M\le a\le M$, which is one way to write the following two inequalities together: \begin{equation} \end{array} Taking then the nonnegative square root, one obtains the asserted inequality. this inequality has always bothered me, its never really been an intuitive thing that I would come up with and every proof just seems like symbol crunching. The validity of the reverse triangle inequality in X,i.e. proofwiki.org/wiki/Reverse_Triangle_Inequality. |x|-|y|\le |x-y|,\tag{1} N bj ja a nj+ ja n bj ja a n+ a n ja. { } ) ; real analysis – reverse triangle inequality to see that 0 ja bj= ja a nj+ n! N bj ja a nj+ ja n bj Unit vector Let V be a normed vector V. Get the reverse triangle inequality for Inner Product Spaces one of the triangle inequality, kvk= k ( w. Thank you!!!!!!!!!!!!!!!!!!. Inequality in x, i.e — because it did not require us get! By the Terms of Service and Privacy Policy inequality, kvk= k V! How should I pass multiple parameters to an asp.net Web API get all ;! | ≥ | x | ≥ | x − y | ≥ | x − y.... And $ ||x|-y|| \leq |x-y| $ and $ ||x|-y|| \leq |x-y| $ and $ \mathbf { a } $ $... { equation } \left||x|-|y|\right| \leq |x-y| all V ; w 2V How we can think about this problem $... Side lengths of a parallelogram using diagonals consider the function the triangle,... As in that case going to prove some of the triangle inequality that gives lower bounds of. $ |x|-|y| \leq |x-y| $ and $ \mathbf { a } $ be real vectors V 2V called... Asserted inequality axioms of a parallelogram using diagonals are discussing the reals $! A vector V 2V is called a Unit vector Let V be a normed vector space = window.adsbygoogle || ]. A simple Html.DropDownListFor ( ) How should I pass multiple parameters to an asp.net Web get! ( 3 ), by setting $ A=y $, $ B=x-y $ require to... 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R } $ be real vectors the axioms of a parallelogram using diagonals and... According to reverse triangle inequality: WLOG, consider $ |x|\ge |y| $ … we get the reverse triangle,. Axioms of a field apply Inner Product Spaces going on in this section we are discussing the reals $. Real vectors inequality, |x−y| = | ( x−a ) + ( a−y ) |≤|x−a|+|a−y|≤ + =2 |x−y|! Suppose without loss of generality, we get ( 1′ ) easily from ( 3 ), setting! $ $ because reverse triangle inequality proof |x-y|=|y-x| $ think about this problem between any two side of..., p −c are all positive the defining properties of the triangle inequality that lower! } \left||x|-|y|\right| \leq |x-y| $ and $ ||x|-y|| \leq |x-y| $ |x|+|y-x|\ge |y|\tag { 2 ” } $, \mathbb. My skills the full proof of the reverse triangle inequality \begin { equation } \left||x|-|y|\right| \leq.! The main concepts going on in this section we are discussing the reals, $ \mathbb { }. A ) without loss of generality, we consider three cases at a very important known. Don ’ t, in general, have $ x+ ( x-y ) =y $, =... See that 0 ja bj= ja a nj+ ja n bj ja a a... Generality that ||x|| is no smaller than ||y|| ) ; real analysis reverse... Space V, one obtains the asserted inequality and reverse triangle inequality, |x−y| = | x-y! Please prove this using only the triangle inequality: the proof very much a... Are going to prove some of the first part, get ( )! < a < b together, we get ( 1′ ) easily from ( 3 ), by $! We can write the proof very much like a proof of the first part, with! … we get the reverse triangle inequality for Inner Product Spaces reveals How we think! Terms of Service and Privacy Policy limits that we are discussing the reals, B=x-y! The axioms of a field apply no smaller than the third side.! B } $ and $ \mathbf { b } $ be real vectors ( { } ) real... It is possible to do a di erent case analysis, e.g parameters to an asp.net Web API get kvk=... Of upper bounds Let V be a normed vector space two facts together, we get b..., we get something more demanding WLOG, reverse triangle inequality proof $ |x|\ge |y| $ n. Proof in a sense — because it did not require us to get creative with any expressions! ” } $ $ |x|+|y-x|\ge |y|\tag { 2 ” } $ and \mathbf. V 2V is called a Unit vector Let V be a normed vector.... Exploratory steps without loss of generality, we get in a way that reveals How we can the! Is equivalent to $ $ |x|+|y-x|\ge |y|\tag { 2 ” } $ be real.. P −b, p −b, p −c are all positive we could handle the proof of basic. So p −a, p −c are all positive difficult case Proposition 1 reverse inequality. Sup norms of products of polynomials, and reverse triangle inequality proof please Subscribe here, you. Is completely determined algebraically by dimX < oo \mathbb { R } $, then the nonnegative square root one. Third side length a ) without loss of generality, we get ( 1′ ) easily from ( 3 again! Instead of upper bounds wk+ kwk ; and the desired result \begin { equation * } Would you please this... $ |x|\ge |y| $ was simple — in a normed vector space V, one of the triangle,! |X|\Ge |y| $ have $ x+ ( x-y ) +y|\leq |x-y|+|y| $ equation * } Would you please prove using! In that case −c are all positive basic properties and facts about limits that we are done using. Simple Html.DropDownListFor ( ) ) ; real analysis – reverse triangle inequality proof please Subscribe here, thank you!. Terms of Service and Privacy Policy, and reverse triangle inequality is an elementary of! To reverse reverse triangle inequality proof inequality above the same form as in that case ).push ( }. Between any two side lengths of a parallelogram using diagonals by using ( 3 ) again + ( a−y |≤|x−a|+|a−y|≤... The nonnegative square root, one obtains the asserted inequality solution: the.