The proof was simple — in a sense — because it did not require us to get creative with any intermediate expressions. cr(X) 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 plane geometry, the statement is: [19] Any side of a triangle is greater than the difference between the other two sides . Let y ≥ 0be ﬁxed and consider the function 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). Then apply $|x| = |(x-y)+y|\leq |x-y|+|y|$. Problem 6. $$1 Young’s inequality: If p,q > 1are such that 1 p + 1 q =1, then xy ≤ xp p + yq q. |-x-y|=x+y\leq-x+y=-(x-y),&y\geq-x\geq0\\ Triangle inequality giv es an upp er bound 2 − , whereas reverse triangle ine qualities give lower bounds 2 − 2 √ 2 for general quantum states and 2 − 2 for classical (or commuting) |x|-|y|\ge -|x-y|\;.\tag{2} These two results mean that i.e. Privacy policy. \bigl||x|-|y|\bigr| PROOF By the triangle inequality, kvk= k(v w) + wk kv wk+ kwk; and the desired conclusion follows. |-x+y|=-x+y=-(x-y),&-x\geq-y\geq0\\$$ Reverse triangle inequality. But wait, (2′) is equivalent to $$The proof of the triangle inequality is virtually identical. a\le M,\quad a\ge -M\;. Hope this helps and please give me feedback, so I can improve my skills. So p −a, p −b, p −c are all positive. Apply THE SQUEEZE THEOREM (Theorem 2.5. proofwiki.org/wiki/Reverse_Triangle_Inequality. \end{equation*} The validity of the reverse triangle inequality in a normed space X. is characterized by the finiteness of what we call the best constant cr(X)associ­ ated with X. \end{equation*} \end{array} We could handle the proof very much like a proof of equality. this inequality has always bothered me, its never really been an intuitive thing that I would come up with and every proof just seems like symbol crunching. Let \mathbf{a} and \mathbf{b} be real vectors. Sas in 7. d(f;g) = max a x b jf(x) g(x)j: This is the continuous equivalent of the sup metric. (b)(Triangle Inequality). =& Then ab 0, so jabj= ab. We get. Thus, we get (1′) easily from (3), by setting A=y, B=x-y. using case 1) x;y 0, and case 2) x 0, y … In this section we are going to prove some of the basic properties and facts about limits that we saw in the Limits chapter. The proof is as follows. Also, … |y|+|x-y|\ge |x|\tag{1′} Hence: (a)Without loss of generality, we consider three cases.$$ The Triangle Inequality for Inner Product Spaces. -|x-y| \leq |x|-|y| \leq |x-y|. Proof. . The proof of the triangle inequality follows the same form as in that case. Proof. |y|-|x| \leq |y-x| Hölder's inequality was first found by Leonard James Rogers (Rogers (1888)), and discovered independently by Hölder (1889) We will now look at a very important theorem known as the triangle inequality for inner product spaces. Thus we have to show that (*) This follows directly from the triangle inequality itself if we write x as x=x-y+y Rewriting $|x|-|y| \leq |x-y|$ and $||x|-y|| \leq |x-y|$. 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