$||x|-|y|| \leq |x-y|$  when $(x,y \in R^k)$  In Principles of MA(Rudin), the author said one sees easily that      $||x|-|y|| \leq |x-y|$  when $(x,y \in R^k)$ (p.88, Rudin)   from the triangle inequality. But I'm not sure how to use the triangle inequality to show this. Can you help me show this?
108965
I'm trying to show that $\lvert \lVert x \rVert - \lVert y \rVert \rvert \overset{\heartsuit}{\leq} \lVert x-y \rVert$. A hint would be nice.
92685
Can you help me with this inequality?  $|x-1| +|x+1| &lt;1$.  I think that it doesn't exist an $x$ that satisfies this inequality, but I'm not really convinced about my answer.
91390
how do you prove this inequality? $|\sqrt[n]{x}-\sqrt[n]{y}| \le \sqrt[n]{|x-y|}$ At first glance I thought the triangle inequality would be useful but it's not form what I see. If I raise everything to the power of n I get the following: Possibility 1 (n uneven) : $|x + ... -y| \le |x-y|$ I guess I would have to prove that ... $\le$ 0 but I don't know how to do that. Possibility 2 (n even) : $|x + ... +y| \le |x-y|$ Same as above but now I have to prove that ... $\le$ -y. Am I even on the right track here? Any help would be appreciated. Thanks in advance.
98279
I have derived the inequality if $k&gt;1$,  ${n(n-1)⋯(n-k+1)\over k!} ({1\over n})^k&lt;{(n+1)n⋯(n-k+2)\over k!} ({1\over n+1})^k$  But, my problem is how to use this inequality to prove that if $n\geq1$, $(1+{1\over n})^n&lt;(1+{1\over n+1})^{n+1}$
106114
pp  and ,   illustrates Triangle Inequality. But without referring to Triangle Inequality at all, how to picture  Reverse Triangle Inequality for all $\mathbf{a,b} \in C^n$? I don't want proofs or rigour.
99514
Can anyone see how to prove the following?  If $x$ and $y$ are real numbers with $y\geq 0$ and $y(y+1) \leq (x+1)^2$ then $y(y-1) \leq x^2$.  It seems it is true at least according to Mathematica.
77694
prove $|z_1 + z_2| \geq ||z_1| - |z_2||$ I have proved the original triangle inequality.  Using the original I have ended up with $$|z_1+z_2| \geq |z_1| - |z_2|$$ and $$|z_1+z_2| \geq -(|z_1|-|z_2|)$$  but does this mean the result holds? I mean, if we had say $|x| \geq 10$, then that means that $x \geq 10$ or $x \leq -10$, so i don't see how to conmclude the result from here
20980
So I was given the task of proving the following variant of the triangle inequality using only the properties of the absolute value:  $\vert\lvert x\rvert -\lvert y \rvert \rvert \leq \lvert x+y\rvert$  Now my initial instinct was to divide it to four cases based on the possible values of x and y, and show that in each case I get a truth expression, but it seems a bit "brute force" to me and I'm not sure this is what my instructor meant. Since I'm not allowed to use the actual triangle inequality and must adhere to absolute value properties, is there any other solution?  (Of course he'll accept this answer if it's correct, but I want to learn what it is I'm missing).
43524
How would I go about proving that  $$|a|+|b| \leq |a+b|+|a-b|$$  Using the triangle inequality?  I tried squaring both sides, yielding:  $$|a|^2 +|b|^2 +2|a||b| \leq 2|a|^2+2|b|^2+2||a|^2-|b|^2|$$  Is it correct to move the terms to the other side? I tried $$2|a||b| \leq |a|^2+|b|^2+2||a|^2-|b|^2|$$ Then it is obvious that  $$0\leq |a|^2+|b|^2-2|a||b| +2||a|^2-|b|^2|$$ $$0\leq (|a|-|b|)^2 +2||a|^2-|b|^2|$$ $$0\leq ||a|-|b||^2 +2||a|^2-|b|^2|$$  But I have neither used the triangle inequality, nor it looks mathematically rigorous for my real analysis class. Any tips and tricks for solving these problems?
122086
I have to prove that $|\arctan x - \arctan y|\leq |x-y|$ for all real numbers $x$ and $y$. I know that it has something to do with the mean value theorem but I just don't know how to solve it. All help is appreciated  Thanks in advance :)
16982
This question looks really simple, but to my (and my co-workers) frustration we were not able to prove this in any way. I tried all triangle formulas known to me but I feel I'm missing the point, and proof will be or much simpler or much more complicated than what I tried.  So, the question:  Given a triangle ABC and point P inside that triangle, prove that for triangle APB the following inequality holds: |AB| + |BC| > |AP| + |PC|  (Actually it doesn't matter for me if it's > or >=).
65484
I know that x > y1 is an inequality but I am in doubt over whether y1 &lt; x &lt; y2 is an inequality or inequalities. Which is right? I think that it may be a math question, but I will be thankful if anyone answers it.
111211
Came across this little practice exercise, and I couldn't properly convince myself of this relation:  Let $X,Y \subset \mathbb{R}^n$ and $g:X\times Y \rightarrow \mathbb{R}$. Show that $$\sup_{y \in Y} \inf_{x \in X}g(x,y) \leq \inf_{x \in X} \sup_{y \in Y} g(x,y).$$  My thinking was starting with $\inf_{x \in X}g(x,\overline{y}) \leq\sup_{y \in Y} g(\overline{x},y)$, but does that even hold for all $\overline{x} \in X$ and $\overline{y} \in Y$?
70120
I am reading Tom Apostol's Analysis and come across this theorem.   Should $a \leq b$ if $a\leq b+\epsilon$ for all $\epsilon &gt;0$?   I don't doubt the proof in the book but I don't understand the intuition or geometric explanation behind this. Could somebody shed some light on this equation? I just started studying analysis on my own.$\ \ $
33224
I need to show \begin{align} \|f*g\|_p \le \|f\|_p\|g\|_1 \end{align} By using the generalized Minkowski inequality instead of just Young's Theorem. I have spent a lot of time, but I keep hitting a dead end. Thanks a million in advance!
20622
I can't seem to work out the inequality $(\sum |x_n|^q)^{1/q} \leq (\sum |x_n|^p)^{1/p}$ for $p \leq q$ (which I'm assuming is the way to go about it).
68258
Let $S$ be the set of all real sequences , for $(x_n) , (y_n) \in S$ , let $d \Big((x_n) , (y_n) \Big) :=0 $ if $x_n=y_n , \forall n \in \mathbb Z^+$ otherwise $d \Big((x_n) , (y_n) \Big) :=\dfrac 1{\min \{n :x_n \ne y_n\}}$ , then is $d$ a metric ?   EDIT : I have only difficulty in proving the triangle inequality