Prove: if   $$a+b+c=\frac1a+\frac1b+\frac1c,\quad0&lt;a,b,c,\in\mathbb R$$   then   $$\frac1{(2a+b+c)^2}+\frac1{(a+2b+c)^2}+\frac1{(a+b+2c)^2}\le\frac3{16}$$   All I tried was $$2a+b+c=a+\frac1a+\frac1b+\frac1c$$ $$LHS=a^2b^2c^2\left(\frac1{((a^2+1)bc+ca+ab)^2}+\frac1{((b^2+1)ca+ab+bc)^2}+\frac1{((c^2+1)ab+bc+ca)^2}\right)$$ that didn't help at all. Thanks.
Let a,b,c be positive real number, such that  $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$. Prove that :  $\frac{1}{(2a+b+c)^2}+\frac{1}{(2b+c+a)^2}+\frac{1}{(2c+a+b)^2} \leq \frac{3}{16}$  Can anyone help me how to deal with it?
How did Lucius Malfoy know Sirius's Animagus form? In Order of the Phoenix, Snape eggs on Sirius at Grimmauld Place:  ‘Tell me, how is Lucius Malfoy these days? I expect he’s delighted his lapdog’s working at Hogwarts, isn’t he?’ [Sirius] ‘Speaking of dogs,’ said Snape softly, ‘did you know that Lucius Malfoy recognised you last time you risked a little jaunt outside? Clever idea, Black, getting yourself seen on a safe station platform ... gave you a cast-iron excuse not to leave your hidey-hole in future, didn’t it?’ Order of the Phoenix - Chapter twenty-four, Occlumency - Page 460 - Bloomsbury  How did Lucius Malfoy know that Sirius was a dog Animagus? How did he know it was Sirius in the train station? Some facts to consider: According to the 's timeline, the characters who are considered to be of the Marauders era (born 1959) were only five years younger than Lucius Malfoy (born 1954), so Sirius and Snape, and the other Marauders-era students, and Lucius Malfoy, would have been at Hogwarts together for two years. Prisoner of Azkaban demonstrates it took Sirius, James, and Peter four to five years to master the Animagus transformation, which would have been at least two years after Lucius Malfoy would have left Hogwarts (Although we don't know when Lucius became a school governor.). A canon-based answer would be great; barring one, a speculative answer in the spirit of canon is welcome. Why is the wild card character * so different between the zip and rm commands? I put together a script to do some file operations for me. I am using the wild card operator * to apply functions to all files of a type, but there is one thing I don't get. I can unzip all files in a folder like this  unzip "*".zip   However, to remove all zip files afterward, I need to do   rm *.zip   That is, it does not want the quotation marks. The unzip, on the other hand, does not work if I just give it the * (gives me a warning that "files were not matched").   Why is this different? To me, this seems like the exact same operation. Or am I using the wild card incorrectly?   in Unix do not really go into this, and I could not locate anything in the rm or zip docs.  I am using the terminal on a Mac (Yosemite). Make apt-get (or aptitude) run with -y but not prompt for replacement of configuration files? When running apt-get -y install &lt;packages ...&gt; on Ubuntu 10.04 I would like apt-get (or aptitude if that makes it easier) to not prompt me when installing additional dependencies (behavior of -y as I understand it) and but not prompt me about overwriting configuration files, instead assume to keep the existing ones always (which is usually the default). Unfortunately --trivial-only seems to be the inverse of -y and not affect the prompt that is shown, according to the man page.  In particular packages auch as samba, nullmailer, localepurge and lighttpd have forced me to interact with the terminal, even though the whole procedure was scripted and meant to be non-interactive. How do I install the latest stable version of LibreOffice? LibreOffice is seeing some dramatic improvements each version and subversion that goes out.   I would like to have always the latest version of LibreOffice on my machine(s). I know of the PPA that exists but is it recommended to use it with 12.04 for daily use? What are the pro and cons of using the PPA? Will it break my system/Libreoffice installation?  Does libreoffice get eventually also upgraded in the Ubuntu repositories or does this upgrade happen only with new releases? Who manages the upgrades of libreoffice in Ubuntu and in the PPA? Best way to find all factors of a given number Apparently,   $$ \sum_{n = 0}^\infty \frac{n}{2^n} $$  converges to 2. I'm trying to figure out why. I've tried viewing it as a geometric series, but it's not quite a geometric series since the numerator increases by 1 every term. Getting Acquaintances to become Friends Started new job. Rollover previous employer 401k to new 401k, IRA or Roth IRA? I recently started a new job and have a 401k from my previous employer that I don't know what to do with.  Should I roll it into my new employer's 401k, a traditional IRA or a Roth IRA? Passive voice phrasing of "The bottle contains two liters" Failed to Login : Invalid session (Try restarting your game) Minecraft Error After restarting my computer when I am presented with the initial login screen (the light gray one with the restart and shut down buttons), I see only two of my three user accounts.  If I log in with one of these accounts, I can see, and log in to, the third (missing) account by either:   Logging off (Apple > Log Out...) to reach the second login screen (the dark gray one with sleep/restart/shut down buttons) Clicking on the fast user switching menu.   How can I restore this account in the initial (light gray) login screen?  I tried repairing disk permissions to no avail.  If I create a new account, it does show up in the list. Improving Review Queues - Design overview II: Changing review bans and other new features On the hardness of data to learn Almost in all texts which are discussing theorems of statistical learning, they assume analyzing arbitrary unknown distribution (the worst case). But in practice different problems (different data) have different levels of hardness, for example linear separable data is easier to learn than the data that are less (or not) separable by hyperplanes.  Are there any works on formalizing the hardness of data (similar to the stuff done in complexity theory)  Note: I am not sure if the same as analyzing VC-dimension or not, but I think it is not. Divisibility Proof with Induction - Stuck on Induction Step I'm working on a problem that's given me the run around for about a weekend. The statement:  For all $m$ greater than or equal to $2$ and for all $n$ greater than or equal to $0$, $m - 1$ divides $m^n - 1$.  Using induction over $n$, the base step comes easy since $m^n - 1$ is $0$ when $n = 0$.  My induction hypothesis is letting $k \geq 0$ and assuming that $m - 1$ divides $m^k - 1$. In order to show that $m - 1$ divides $m^{k+1} - 1$, I obviously need to use the induction hypothesis. However, no matter where I try to use the fact that $m^k - 1 = (m-1)a$ for some $a$ in the integers, the expression $m^{k+1} - 1$ always becomes more difficult to get to $m^{k+1} - 1$ being equal to $(m-1)b$ for some $b$ in the integers.   In other words, I can't figure out any actually helpful way to apply the induction hypothesis with the goal of proving the next step.  Any help would be appreciated! G acts on X transitively, then there exists some element that does not have any fixed points Let $X$ be a transitive $G$-set. ($G$ acts on  $X$ transitively.) If $X$ is finite and has at least two elements, show that there is some element $g$ $\in$ G which does not have any fixed points; that is, such that $g$$($$x$$)$ $\ne$ $x$ for all $x$ $\in$ $X$  I am trying to use contradiction, but it is not very clear to me why this is true. I tried to install Ubuntu 18.10 on my XPS 15 9570 earlier. Everything was working fine until I got to the partition selection part of the installation. That's when the installer crashed and I had to shut down my machine. I think it's because I had my SATA configuration set to RAID ON instead of AHCI, which is now fixed.  Now when I try to run the installer from my bootable USB I get the following error  Failed to open \EFI\BOOT\mmx64.efi - Not Found Failed to load image \EFI\BOOT\mmx64.efi: Not Found Failed to start MokManager: Not Fond Something has gone seriously wrong: import_mok_state() failed   Hoping someone might have some idea as to what is going on How to find the determinant of this $3 \times 3$ Hankel matrix? Today, at my linear algebra exam, there was this question that I couldn't solve.   Prove that $$\det \begin{bmatrix}  n^{2} &amp; (n+1)^{2} &amp;(n+2)^{2} \\  (n+1)^{2} &amp;(n+2)^{2}  &amp; (n+3)^{2}\\  (n+2)^{2} &amp; (n+3)^{2} &amp; (n+4)^{2} \end{bmatrix} = -8$$   Clearly, calculating the determinant, with the matrix as it is, wasn't the right way. The calculations went on and on. But I couldn't think of any other way to solve it. Is there any way to simplify $A$, so as to calculate the determinant? Correct ssh config file settings to tunnel to a 3rd machine I'm trying to tunnel to a server via a bridge server. So far, I've been able to get it working from the command shell properly using the following command:  ssh -A -t bridge_userid@bridgemachine.com ssh -A remote_userid@remoteserver.com   But I've been trying to wrap this into my ~/.ssh/config file and I have troubles. I've tried:  Host axp   User          remote_userid   HostName      remoteserver.com   IdentityFile  ~/.ssh/id_rsa.eric   ProxyCommand ssh -A -t bridge_userid@bridge_userid.com ssh -A remote_userid@%h   But when I do, I get the following error message from remoteserver.com and I'm not sure what is causing it:     ksh: SSH-2.0-OpenSSH_6.8^M:  not found   I know that when I log into remoteserver.com , my shell is /usr/bin/ksh.  I've tried to add path arguments to the ssh commands in the config file, but it made no difference.  Any ideas what it can be? Consecutively Adding Sines One thing, I'm not a mathematician so please be patient. I am still in Algebra II Trig. Leading with that, why does  $$ x_0 = \sin 1, \space  x_1 = x_0 + \sin x_0, \space x_2 = x_1 + \sin x_1 ... $$ and after a while, $$ x = \pi $$  I know this to be true because I have evaluated this on my TI-84 and more deeply evaluated it with this program I made:   To $100000$ Iterations (which I believe is enough)   So what's the math behind it? Why does adding $\sin 1$ in this manner produce $\pi$? Why doesn't doing the same thing using $100$ produce $\pi$? For example,  $$\sin(1) = 0.841470...,\space \sin(1) + \sin(0.8414...), \sin(1) + \sin(0.841470...) + \sin(1.587095126...),\space ... $$ $$ = \pi$$  My question is different from the mentioned because my function is not taking the sin of a sin consecutively, but is this following function: $f(x) = x + \sin x$ and is not $f(x) = \sin x$ and therefore can exhibit different properties I need explained. 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