I have reset my world spawn to a distant place. how can I get back to the original world spawn area?
Finding my house in minecraft
Python Flask shutdown event handler Prove that $a^3+b^3+c^3 \geq a^2b+b^2c+c^2a$ Let $a,b,c$ be positive real numbers. Prove that $a^3+b^3+c^3\geq a^2b+b^2c+c^2a$.   My (strange) proof:  $$ \begin{align*} a^3+b^3+c^3 &amp;\geq a^2b+b^2c+c^2a\\ \sum\limits_{a,b,c} a^3 &amp;\geq \sum\limits_{a,b,c} a^2b\\ \sum\limits_{a,b,c} a^2 &amp;\geq \sum\limits_{a,b,c} ab\\ a^2+b^2+c^2 &amp;\geq ab+bc+ca\\ 2a^2+2b^2+2c^2-2ab-2bc-2ca &amp;\geq 0\\ \left( a-b \right)^2 + \left( b-c \right)^2 + \left( c-a \right)^2 &amp;\geq 0 \end{align*} $$  Which is obviously true.    However, this is not a valid proof, is it? Because I could just as well have divided by $a^2$ rather than $a$:  $$ \begin{align*} \sum\limits_{a,b,c} a^3 &amp;\geq \sum\limits_{a,b,c} a^2b\\ \sum\limits_{a,b,c} a &amp;\geq \sum\limits_{a,b,c} b\\ a+b+c &amp;\geq a+b+c \end{align*} $$  Which is true, but it would imply that equality always holds, which is obviously false. So why can't I just divide in a cycling sum?  Edit: Please don't help me with the original inequality, I'll figure it out. C++ catching all exceptions When at home I use my 13" MacBook Pro hooked up to an 27" external monitor. When working on my 27" screen I prefer to have a visible dock at all times. I have enough screen real estate to afford this luxury.  When I'm on the road working on my 13" screen I prefer to automatically hide the dock as it really gets in the way when screen real estate is limited.  I now manually switch between enabling and disabling hiding (using ⌥⌘D) but I would prefer this to happen automatically. Can I enable hiding the dock on my MacBook monitor and disable hiding the dock on my external monitor? I am looking to build a camera trap for wildlife viewing, able to capture pictures (or even videos) in decent quality to be able to identify distinct features. I do however need to be able to see the pictures in real time, so the camera would need some sort of wireless connection to a receiving device such as a phone.  I thought of using an FPV setup as used in drone racing, but the quality and range seems to be quite low, and the strengths of an analog signal seem to be latency and fps, which are both not important to me. It would be nice to have some sort of digital signal to be able to do some editing, such as storing pictures. It would also be cool to connect several cameras to one device.  Are there any solutions to this problem? Would a raspberry pi/arduino setup work in some way? Thanks! What iron-clad reason did Dumbledore have for believing Snape was on his side "I have never said" versus "I never said" Does a MCMC fulfilling detailed balance yields a stationary distribution? I guess I understand the equation of the detailed balance condition, which states that for transition probability $q$ and stationary distribution $\pi$, a Markov Chain satisfies detailed balance if $$q(x|y)\pi(y)=q(y|x)\pi(x),$$  this makes more sense to me if I restate it as:   $$\frac{q(x|y)}{q(y|x)}= \frac{\pi(x)}{\pi(y)}. $$  Basically, the probability of transition from state $x$ to state $y$ should be proportional to the ratio of their probability densities. Will taking a gap year after my Master's degree improve or degrade my potential of doing a PhD? "Elder brother" or "older brother"? I've read both forms in newspapers and online news: elder brother and older brother. What's the difference between them? When should I use which? Is it possible to create bootable linux on USB drive using windows command line? In Python, you can do this:  import webbrowser webbrowser.open_new("http://example.com/")   It will open the passed in url in the default browser  Is there a ruby equivalent? Make back/forward work on paginated lists Why does a bigger sensor lead to a shallower depth of field? It has been mentioned at    that a larger sensor results in images with a shallower depth of field. Example image:          I understand how sensor size would relate to, for example, field of view, but the relationship with depth of field does not seem straightforward.  It actually seems contradictory - I have more wells on my sensor, and I am able to focus on less number of points.  What is the reason for this effect? Skype window problem in Ubuntu 14.04 How do I correctly clone a JavaScript object? I have an object x. I'd like to copy it as object y, such that changes to y do not modify x. I realized that copying objects derived from built-in JavaScript objects will result in extra, unwanted properties. This isn't a problem, since I'm copying one of my own literal-constructed objects.  How do I correctly clone a JavaScript object? seem to be well-regarded for their versatility and accuracy of metering.  What puzzles me about them is how can they account for the different light transmittance of different lenses? It's easy for an in-camera metering system that meters  so any peculiarity in transmittance is already applied to light when it reaches the metering sensor. But a standalone meter does not have the privilege of looking at the world through the lens it meters for. I tried looking it up in  and , but this issue was not touched upon at all.  This leaves several variants how it could be handled -   light meter uses  for aperture display/input, and photographer is expected to perform the adjustments to and from  (which are used by cameras); light meter assumes some median transmittance of lenses of its era, and automatically accounts for that; photographers simply ignore the matter - even a 75% transmitting lens will be off only by about half a stop, no big deal; each light meter used is expected to know that you have to dial in some exposure compensation to account for the light loss; ...   So how does it work in real world? Let $\psi: C([0,1]; \mathbb{R}) \mapsto C([0,1]; \mathbb{R})  $, be such that if $x \in [0,1]$ then $$\psi(f)(x) = \int\limits_{0}^{x} f(t) dt $$  I am asked to show that $\psi$ is not a contraction but $\psi^{2}$ is. And then conclude that $\psi$ has a unique fixed point.   I´ve managed to prove a statment that helps me conclude the last part but I´m having trouble proving that the function is not a contraction. This is what I've got.   If $f,g \in  C([0,1]; \mathbb{R})$  then $$ |\psi (f) - \psi(g)| \leq \int\limits_{0}^{x} |f(t)-g(t)| dt \leq \sup\limits_{x\in [0,1]} |f(x)-g(x)| \int\limits_{0}^{x} 1 dt = d(f,g) x  $$ Then $|\psi (f) - \psi(g)| \leq d(f,g) x $,  and taking supremum on the left hand side allows me to conclude that $d(\psi (f),\psi (g)) \leq d(f,g) x$  Is there any problem with this procedure? Does anybody has an insight on how to procede with the $\psi^{2}$ part?   Many Thanks! I am using \if condition in a custom command in my LaTeX document. I would like to know where I can find some good explanation about \if command. I followed a  but it's just an example; there are not any explanation. How to close entire application when application has more than 2 forms without showing messagebox more than once