How do you derive that  | |x|-|y| |&lt;= |x-y|
8964
I've seen the full proof of the Triangle Inequality  \begin{equation*} |x+y|\le|x|+|y|.  \end{equation*} However, I haven't seen the proof of the reverse triangle inequality:  \begin{equation*} ||x|-|y||\le|x-y|. \end{equation*} Would you please prove this using only the Triangle Inequality above?  Thank you very much.
106291
Let $x,y\in \left( 0;1\right).$ I want to prove that  $$\left( \dfrac {x} {y}\right) ^{x}\left( \dfrac {1-x} {1-y}\right) ^{1-x}\geqslant 1$$
10198
Prove if $|z| &lt; 1$ and  $ |w| &lt; 1$, then $|1-zw^*| \neq 0$ and $| {{z-w} \over {1-zw^*}}| &lt; 1$Given that $|1-zw^*|^2 - |z-w|^2 = (1-|z|^2)(1-|w|^2)$I think the first part can be proven by saying $|1-zw^*| = 0$ if and only if $zw^*$ = 1.  And given the conditions that cannot be true. However I don't know if this part is right.
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 $\overline{y} \in Y$?
122977
I have to prove that the result of $x * y \in G$ so $\frac{x+y}{1+xy} \in (-1, 1)$. So $x &gt; -1$ and $y &gt; -1$ at the same time $x &lt; 1$ and $y &lt; 1$.  If I multiply the first 2 expressions I obtain $xy &lt; 1$ which is true, and if I sum up the last 2 I get $x + y &lt; 2$. At this point I am not sure if I am doing well. How should I continue this problem?
76208
So if $a &lt; b &lt; c &lt; d$ , how do I express the set $[a,b] \cup [c,d]$ as the difference of two intervals? Is it as simple as $[a,d] -[b,c]$?
92174
If $x,y\in\mathbb R$, I have problems to show that  $$\lfloor x\rfloor+\lfloor y\rfloor\le \lfloor x+y\rfloor\le \lfloor x\rfloor+\lfloor y\rfloor + 1 $$  Can someone help me?
30991
If $x^{2}-x\in Z(R)$ for all $x\in R$, then $R$ is commutative.  I need to proof this theorem and I have something like this below. However, I do not know how to continue this proof.   $(x+y)^{2}-(x+y)=x^{2}+xy+yx+y^{2}-x-y=(x^{2}-x)+(y^{2}-y)+xy+yx\in Z(R)$  so $xy+yx\in Z(R)$
7044
I understand intuitively that this is true, but I'm embarrassed to say I'm having a hard time constructing a rigorous proof that $|a+b| \leq |a|+|b|$. Any help would be appreciated :)
78884
I am trying to simplify $E[YE[Y|X]|X]$ can I use this property:   $$E[E[Y|X]|X]= E[Y|X]$$  If yes I have never seen a Proof of this property (that seems very reasonable), could I have a reference? If I am right the correct result would be $$E[YE[Y|X]|X] = E[(Y|X)]^2$$
7683
Let $V$ and $W$ be vector spaces over a field $F$. Let $L(V,W) = \{T:V\to W : T \text{ is linear} \}$, that is, $L(V,W)$ is the collection of all linear functions from $V$ to $W$. For $S,T \in L(V,W)$ and $a \in F$, define new functions $(S+T):V \to W$ and $ aT :V \to W$ by $(S+T)(x) = S(x)+T(x)$ ; $(aT)(x) = a(T(x))$.   Prove that $S+T$ belongs to $L(V,W)$ and $aT$ belongs to $L(V,W)$.
2419
Prove $||a|-|b|| \leq |a-b|$  I do it by making 4 cases :  Case 1 : $ a \geq 0$, $ b\geq 0$  So we have   $||a||-||b||  = |a - b|$ . hence proved  Case 2 :   $a &lt; 0 , b&lt; 0$  $|-a+b| = |-(a-b)|=|a-b|$  Case 3  $a &gt;0, b&lt;0$  $||a||-||b|| =|a+b| \leq |a|+|b| \leq a -b \leq |a-b| $  Case 4:  $a &lt;0 ,b&gt;0$  $|-a+b| = |-(a-b)| = |a-b|$  Is this OK ? Thanks
43870
Prove:$|x-1|+|x-2|+|x-3|+\cdots+|x-n|\geq n-1$   example1: $|x-1|+|x-2|\geq 1$  my solution：(substitution)  $x-1=t,x-2=t-1,|t|+|t-1|\geq 1,|t-1|\geq 1-|t|,$  square,  $t^2-2t+1\geq 1-2|t|+t^2,\text{Since} -t\leq -|t|,$  so proved.  question1 : Is my proof right? Alternatives?  one reference answer:   $1-|x-1|\leq |1-(x-1)|=|1-x+1|=|x-2|$  question2 : prove:  $|x-1|+|x-2|+|x-3|\geq 2$  So I guess:( I think there is a name about this, what's that? wiki item?)  $|x-1|+|x-2|+|x-3|+\cdots+|x-n|\geq n-1$  How to prove this? This is  question3. I doubt whether the two methods I used above may suit for this general case.  Of course, welcome any interesting answers and good comments.
86572
How should I approach obvious proofs like these    I have been trying but couldn't find an elegant way to work these. Any help is highly appreciated ! Especially looking for a nice proof/hint for first problem
71323
Prove: $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}$$ for all x, y positive   $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}-\sqrt{x}-\sqrt{y}\geq 0$$  $$\frac{x\sqrt{x}+y\sqrt{y}-x\sqrt{y}-y\sqrt{x}}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{x(\sqrt{x}-\sqrt{y})+y(\sqrt{y}-\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{x(\sqrt{x}-\sqrt{y})-y(-\sqrt{y}+\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{(x-y)(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{(\sqrt{x}-\sqrt{y})^2(\sqrt{x}+\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$  All the elements are positive and if $\sqrt{x}=\sqrt{y}$ we get $0$  Is the proof valid?
20557
Given $\exists y. \forall x. p(x,y)$, use Fitch-style natural deduction system to prove $\forall x.\exists y.p(x,y)$.  I know this question has been asked , but based on that answer I'm not able to move forward from step 3 below. I'm using  as a proof assistant by the way, maybe that's the issue. Any help is very much appreciated.  1.  EY:AX:p(X,Y)     Premise 2.  AX:p(X,Y).       Assumption 3.  p(X,Y).          Universal Elimination: 2
106369
Prove that for $x&gt;0$  and $0&lt;p&lt;q$:  $$\left( 1+ \frac{x}{p} \right) ^{p}  &lt; \left( 1+ \frac{x}{q} \right)  ^{q} $$