I want to prove that, $$\left| x-y\right|\geq\left| x\right| -\left| y\right|$$ I approached the problem like this af:  Since $-\left|z\right|\leq z\leq\left|z\right|$ is true for any real number $z$, we can write the same for reals $x$ and $y$: $$-\left|x\right|\leq x\leq\left|x\right|$$ $$-\left|y\right|\leq y\leq\left|y\right|$$ We can subtract one expression from the other and get: $$-(\left|x\right|-\left|y\right|)\leq x-y\leq\left|x\right|-\left|y\right|$$  And, since $-a\leq z\leq a$ is equivalent to $\left|z\right|\leq a$ (for any real number $z$), we can write the following: $$\left| x-y\right|\leq\left|x\right|-\left|y\right|$$  This is, in fact, the opposite of what I intended to prove. I know the correct derivation though. Since $-\left|z\right|\leq z\leq\left|z\right|$ is true for any real number $z$, we can write the same for $y$ and $x-y$: $$-\left|y\right|\leq y\leq\left|y\right|$$ $$-\left|x-y\right|\leq x-y\leq\left|x-y\right|$$ By adding the two inequalities, we get: $$-(\left|x-y\right|+\left|y\right|)\leq x\leq\left|x-y\right|+\left|y\right|$$ And, since $-a\leq z\leq a$ is equivalent to $\left|z\right|\leq a$ (for any real number $z$), we can write the following: $$\left| x\right|\leq\left|x-y\right|+\left|y\right|$$ Finally, we can subtract $\left|y\right|$ from both sides and voila: $$\left| x\right|-\left| y\right|\leq\left|x-y\right|$$  This equation is simply failsafe, it holds true for all real values of $x$ and $y$. I'm more curious about $\left| x-y\right|\leq\left|x\right|-\left|y\right|$. It seems algebraically correct but ho, it's experimentally BS. Try placing $x=-6$ and $y=-7$ for instance. The values for $\left| x-y\right|$ and $\left|x\right|-\left|y\right|$ are respectively $1$ and $-1$.   Total BS, no?  So where did I screw up? I must have screwed up algebraically, but where? What's so wrong about $\left| x-y\right|\leq\left|x\right|-\left|y\right|$, algebraically?
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.
85884
Show that for $x,y,z &gt; 0$ the inequality is true: $\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+x+y+z \geq \frac{(x+y)^2}{y+z}+\frac{(y+z)^2}{z+x}+\frac{(z+x)^2}{x+y}$  I have tried Holder, but i had no luck. Please give me a hint of how to start.
31053
For any $\epsilon&gt;0$ and $a,b\in\mathbb{R}$ show that $$|a-b|&lt;\epsilon\Rightarrow  |a|&lt;|b|+\epsilon$$   In some notes I found the following property $$|a|-|b|\leq ||a|-|b||\leq|a-b|$$ So $$|a-b|&lt;\epsilon\Rightarrow |a|-|b|&lt;\epsilon\Rightarrow |a|&lt;|b|+\epsilon$$  I have two doubts  1) Is it right?  2)In triangular inequality I have that $$|a+b|\leq |a|+|b|$$ $$|a+(-b)|\leq |a|+|-b|\leq|a|+|b|$$  Is wrong do that?
79801
I have to do this exercise for my math study, and I'm having trouble with doing the second part of it.  Let $x, y, \epsilon \in \mathbb{R}$ and $\epsilon &gt; 0$. Prove: $|x - y| \leqslant \epsilon$ $\forall \epsilon &gt; 0 \Leftrightarrow x = y$  I think I have the left implication $\Leftarrow$:  Assume $x = y \Rightarrow x - y = 0 \Rightarrow|x - y| = 0 = |0| \Rightarrow|x - y| = 0 &lt; \epsilon \Rightarrow |x - y| \leqslant \epsilon$ $\forall \epsilon &gt; 0 \in \mathbb{R}$  Is this argument correct?  For the right implication, I do only have an idea of how to prove it. I think I have to assume first that $x &gt; y$ and then $x &lt; y$ and get a contradiction form both.   Could you please explain me the right implication and tell me if my left implication is correct?  Thanks in advance!
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
22053
The formula we need to prove is $$ \displaystyle \iint_{x^{2}+y^{2} \leq 1} e^{x^{2}+y^{2}} d x d y \leq\left[\int_{-\frac{\sqrt{\pi}}{2}}^{\frac{\sqrt{\pi}}{2}} e^{x^{2}} d x\right]^{2}  $$ As I already done the left side $$ \iint_{x^{2}+y^{2} \leq 1} e^{x^{2}+y^{2}} d x d y  = \int_{-\pi}^{\pi} \mathrm{d}\varphi \int_{0}^{1} e^{r^2}r\mathrm{d}r  = 2\pi \int_{0}^{1} e^{r^2}r\mathrm{d}r = \left.\pi e^{r^2}\right|_{0}^{1} = \pi(e-1) $$ I still have no idea about the right side.
110459
This problem is from Romanian G.M. and although it is very short, it is also (I think) very hard. Let $x, y, z$ real non-negative numbers such that $x+y+z=3$. Prove that   $$27 \leq (x^2+2)(y^2+2)(z^2+2) \leq 44 $$  I didn't succeed in any of the inequalities, I only managed to find the equality cases ($x=y=z=1$ for the first one and one of them $3$ and the others $0$). Any hint/idea/solution is welcome.
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 :)
28521
If $x,y,z\in \mathbb{R}\;,$ Then Range of $$\frac{|x+y|}{|x|+|y|}+\frac{|y+z|}{|y|+|z|}+\frac{|z+x|}{|z|+|x|}\,$$    $\bf{My\; Try::}$ Here $x,y,z$ Not all Zero Simultaneously.   Now Using $\bf{\triangle\;  Inequality}\;,$ We get $|x+y|\leq |x|+|y|$  Similarly $|y+z|\leq |y|+|z|$ and $|z+x|\leq |z|+|x|$  So we get $$\frac{|x+y|}{|x|+|y|}+\frac{|y+z|}{|y|+|z|}+\frac{|z+x|}{|z|+|x|}\leq 3$$  Now I did not understand How can I calculate $\min$ of that expression, Help Required  Thanks
6481
I would appreciate if somebody could help me with the following problem  Let $|x|\leq1, |y|\leq 1$. Show that  $$ 0  \leq x^{2}+y^{2} -2x^{2} y^{2}+2xy  \sqrt{1-x^{2}}  \sqrt{1-y^{2}} \leq 1 $$
25116
The question is to show that for any $x,y \in\Re^n$  $||x+y||\cdot||x-y||\le||x||^2+||y||^2$  with equality if and only if $\langle x,y \rangle= 0.$  I have that $||x+y||^2=||x||^2+2 \langle x,y \rangle+||y||^2$ , which helps me prove equality but I'm lost on how to show this and show that the inequality part.
33562
I can show that for $x &gt; 0$ and $r_{i} &gt; 0$ we have $$ \left(\, x + r_{1}\,\right)\ldots\left(\, x + r_{n}\,\right)\ \geq\ \left[\, x + \left(\, r_{1}\ldots r_{n}\,\right)^{1/n}\,\right]^{n}.$$ However, I can't do this using straight up induction, strong or weak. Can someone do this?   You can show the inequality holds for $n$ if it holds for $n+1$ using $r_{n+1} = (r_1 \cdots r_n)^{1/n}$, so going down isn't a problem, but going up has eluded me. Additionally, you can show that if it holds for $n=a$ and $n=b$ then it holds for $n=ab$. Using powers of two and the above is sort of a way to prove the above by induction but it certainly isn't "normal".  You can also prove the above using Lagrange multipliers (which isn't surprising). The last proof I know is where you compare coefficients of $x^k$ and use the AM-GM inequality.
91951
Any hints on how I can prove the inequality:     $$|\sqrt{x}-\sqrt{y}| \leq \sqrt{|x-y|},\quad x,y \geq 0$$   Thank you.
77867
Let $a$ and $b$ be positive numbers, and $n \in \mathbb{N}$. Prove that (using Rearrangement Inequality) $$(n+1)\left(a^{n+1}+b^{n+1}\right) \geq (a+b)\left(a^n+a^{n-1}b+\ldots+b^{n}\right). $$  Thanks :)
35514
Let $f(x)$ be a continuous function. Prove that $\left|f(x)\right|$ is also continuous.  Is it correct to say that, by the reverse triangle inequality, $\left|f(x)-f(c)\right| \geq \left|f(x)\right|-\left|f(c)\right|$ in all cases, so we will always have that for all $\left|x-c\right|&lt;\delta$ implies $\left|f(x)-f(c)\right|\leq \epsilon$?  I am not sure if my solution adequately uses the functional form, since I just used the equation and not my actual function. Please help with the proper solution!
11077
Prove the inequalities $|z_1 +z_2| \geq \frac{1}{2}(|z_1|+|z_2|)|\frac {z_1} {|z_1|} +\frac {z_2} {|z_2|}|$  Here is what I got so far  I start with $z_1=r_1 cis \theta_1$ (at this time we are not allowed to use $e^{i\theta}$ but $cis \theta =e^{i \theta}$)  $z_2 =r_2 cis \theta_2$. I will focus on the right hand side of the inequality   $RHS= \frac{1}{2}(r_1 +r_2)| \frac {r_1 cis \theta_1}{r_1} +\frac{r_2 cis \theta_2}{r_2}|=\frac{1}{2}(r_1 +r_2)|cis \theta_1 +cis \theta_2|=\frac{1}{2}(r_1 +r_2)((\cos \theta_1 +\cos \theta_2)^2 +(\sin \theta_1+\sin \theta_2)^2 )^{1/2}=\frac{1}{2}(r_1 +r_2)(2+2(cos \theta_1 cos \theta_2+ sin \theta_1 sin\theta_2))^{1/2}=\frac{1}{2}(r_1 +r_2)(2+2(cos (\theta_1 -\theta_2))^{1/2}=(r_1+r_2)(1+\cos(\theta_1-\theta_2))$  but I'm not sure how this result can help me prove the inequality, because I can't see how the left hand side related to this.
31604
Let $a,\ b,\ c$ be positive real numbers. Prove that: $$\left(a+\dfrac{bc}{a}\right)\left(b+\dfrac{ca}{b}\right)\left(c+\dfrac{ab}{c}\right)\geq 4\sqrt[3]{\left(a^{3}+b^{3}\right)\left(b^{3}+c^{3}\right)\left(c^{3}+a^{3}\right)}\tag1$$