How do you prove $|y|&lt;x \Leftrightarrow -x&lt;y&lt;x$ I tried using the fact that $|a|=\sqrt{a^{2}}$ but it got messy.how would I go about proving this
122936
I know I have to prove both sides here so: $\implies$: Suppose $|a| &lt; b$. Since $|a|$ is a positive number, we know $b$ is a positive number greater than $|a|$:  If $a$ is positive, $a &lt; b$ If $a$ is negative, $a &lt; b$  But for $|a| &lt; b$, it has to be that $a &gt; -b$. $a &lt; b$ and $a &gt; -b$, therefore $–b &lt; a &lt; b$ is true. $\impliedby$: Suppose $-b &lt; a &lt; b$.  If $a$ is positive, then $b$ is a greater positive number, and $-b$ would be less than $a$ If $a$ is negative, then $b$ can only be a greater positive number for $-b &lt; a$ to hold  So, $b$ is a greater positive integer than $a$. Therefore, it must be that $|a| &lt; b$. This was what I came up with but it seems convoluted and kind of forced with the cases. Any suggestions?
69535
Let $f(x) = |x|$ for $x \in [-1,1]$ and extend $f$ to all of $\mathbb R$ by requiring $f(x+2)=f(x)$. Prove that $|f(x)-f(y)|\le|x-y|$ for all $x$ and $y$.   This is easy to observe on a graph but my proof is long and a bit tedious. Just curious if there is a quick/elegant way?
122604
How can I prove      $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}...}}}}=2$$   I don't know which method can be used for this?
59700
If $x,y\in \mathbb{R}$, and $x&gt;y$, how to show $(x-y)^{((x-y)/(2x-y))}\times (x+y)^{((x)/(2x-y))}&gt;x$?  I know I have to use AM-GM inequality, but it is not clear how.
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.
113746
Show that for $u,v \in \mathbb{C}$ with $|u|&lt;1, |v|&lt;1$, and $\bar{u}v\neq u\bar{v}$, we always have $$\left|\left(1+|u|^2\right)v-\left(1+|v|^2\right)u\right|&gt;\left|u\bar{v}-\bar{u}v\right|.$$
30189
Can you help me to prove that $$|x-y|\leq |x|+|y|$$ I get a proff of this equality, but it's very short and I don't know if it's correct.
48128
Verify that  $\sqrt{2}\left\| z \right\| \ge \left|\Re(z)\right| + \left|\Im(z)\right|.$  I started off noting that $z=x+iy$ and that $Re(z)=x$ and $Im(z)=y$  Then I know that I have to square both sides, giving me   $$2(x^2 +y^2) \ge x^2+2|xy|+y^2$$  I'm not sure what to do after this.
64286
I am trying to prove,  $\lim_{x \to a} \sqrt x = \sqrt a$  As per the definition of limit for every $\epsilon &gt; 0$ there is some $\delta &gt; 0$ such that   $ 0 &lt; |x-a| &lt; \delta$ implies $|f(x) - \sqrt a| &lt; \epsilon$  $|\sqrt x - \sqrt a|  = \frac {|x -a|}{\sqrt x + \sqrt a} $  since  $ 0 &lt; |x-a| &lt; \delta$  $\frac {|x -a|}{\sqrt x + \sqrt a} &lt; \delta$  Can I stop my proof at this point since I found $\epsilon$ ( which is $\delta$ in this case)
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?
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.
31604
Let $a,\ b,\ c$ be positive real numbers. Prove that: $$\left(a+\dfrac{bc}{a}\right)\left(b+\dfrac{ca}{b}\right)\left(c+\dfrac{ab}{c}\right)\geq 4\sqrt[3]{\left(a^{3}+b^{3}\right)\left(b^{3}+c^{3}\right)\left(c^{3}+a^{3}\right)}\tag1$$
121264
I want to prove that for each real number $x&gt;0$, the statement  $${\left(\frac{x+1 } {2}\right)}^{\!x+1} \leq x^x$$  is correct.   I’ve tried using mathematical induction methods, but haven’t found a solution. Although, using any method is appreciated.
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
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?
2180
Show that $|x-y|\leq |x|+|y|$.   I know this probably applies the triangle inequality in some way, but I can't figure it out.
106369
Prove that for $x&gt;0$  and $0&lt;p&lt;q$:  $$\left( 1+ \frac{x}{p} \right) ^{p}  &lt; \left( 1+ \frac{x}{q} \right)  ^{q} $$