If $a,b&gt;0$ and $a+b=1\;,$ Then minumum value of $\displaystyle \left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2$ is   $\bf{My\; Try::}$ Let $a=\sin^2 \theta$ and $b=\cos^2 \theta\;,$ Then We have to minimize   $$\displaystyle f(\theta) = (\sin^2 \theta+\csc^2 \theta)^2+(\cos^2 \theta+\sec ^2 \theta) = \sin^4 \theta+\csc^4 \theta+\cos^4 \theta+\sec^4 \theta+4$$  So $$f(\theta) = 1-2\sin^2 \theta\cos^2 \theta+\frac{1}{\sin^2 \theta \cos^2 \theta}+4$$  So $$f(\theta) = 1-\frac{\sin^2 2\theta}{2}+\frac{4}{\sin^2 \theta}+4$$  Now Put $\sin^2 2\theta = t\;, t\in \left(0,1\right]$  So we get $$f(t)=5-\frac{t}{2}+\frac{4}{t}\;,$$ So we get $\displaystyle f'(t) = -\frac{1}{2}-\frac{4}{t^2}&lt;0\;\forall t \in (0,1]$  So $$f(1)_{\min} = 5-\frac{1}{2}+4=9-\frac{1}{2}=\frac{17}{2}$$  But answer given as $$\frac{25}{2}$$  plz help me, Where I am wrong, Thanks
If $a, b$ are positive real numbers and $a+b = 1$, prove that :  $$\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 \geq \frac{25}{2}$$  I can see that the value $\frac{25}2$ is attained for $a=b=\frac12$. But I do not know how to show that this is the minimal possible value.  Thank you.
Earlier today, I was suspended from chat due to an inappropriate message (I totally deserved it, but that's not the point). I looked at my phone, and this is all that it told me:    ...which confused me because chat was still moving as normal. I didn't realize that I had been suspended until I checked on my computer to see what the heck was going on.  To me, "This room is read-only" sounds like some sort of database lockup or something related specifically to the room, not the user. It doesn't give any feedback to indicate that the room will be "read-only" for 30 minutes, and doesn't mention that all other rooms are also "read-only". Even a message saying just "You have been suspended from chat" would be more helpful. I have a 1TB FAILING hard drive, a brand new 1 TB hard drive, a bootable ubuntu usb, and a working computer.  All I need is a step by step process of how to safely clone as much data as possible from one to the other. NOTE: THE DRIVE IS FAILING.  I know I need to enable multiverse and then install and use ddrescue to complete the task. But everywhere I look online, I can't get a straight path to the task. I know that this can be all done within the terminal and that's what I seek.  How can I do this? I only get four video output options Repost it with a minimal working example for the amsmath must be loaded earlier than mathspec problem See the following mwe:  \documentclass{article} \usepackage{amsmath} \usepackage{mathspec} \usepackage{unicode-math}  \begin{document} \begin{equation*}   1 + 2 = 3  \end{equation*} \end{document}   Run the above code with xelatex will meet the following error:  ! Package mathspec Error: `amsmath' must be loaded earlier than `mathspec'.  See the mathspec package documentation for explanation. Type  H &lt;return&gt;  for immediate help.  ...                                                l.54 \RequirePackage{amsmath}  ?    $ xelatex --version XeTeX 3.14159265-2.6-0.999991 (TeX Live 2019)  Regards Transaction log reported full but is not How do I determine if my integrated Intel graphics support "Shader Model 3"? I am trying to play a Steam game on Ubuntu.  The game requires "Graphics: Shader model 3".  How do I determine if my integrated Intel graphics support Shader Model 3?  At present, when I launch the game, I hear the audio, but the screen is entirely black.  Running glxinfo | grep OpenGL yields:  OpenGL vendor string: Intel Open Source Technology Center OpenGL renderer string: Mesa DRI Intel(R) Haswell Mobile  OpenGL core profile version string: 4.5 (Core Profile) Mesa 18.2.2 OpenGL core profile shading language version string: 4.50 OpenGL core profile context flags: (none) OpenGL core profile profile mask: core profile OpenGL core profile extensions: OpenGL version string: 3.0 Mesa 18.2.2 OpenGL shading language version string: 1.30 OpenGL context flags: (none) OpenGL extensions: OpenGL ES profile version string: OpenGL ES 3.1 Mesa 18.2.2 OpenGL ES profile shading language version string: OpenGL ES GLSL ES 3.10 OpenGL ES profile extensions:   I am not sure which (if any) of the above lines answer my question. I have been wondering why anyone ever thought that we could find an expansion for eigenstates of some perturbed Hamiltonian in terms fo those for the basis Hamiltonian. My lecturer insisted that this was because the eigenstates of the Hamiltonian are complete', however from my (limited) understanding of the mathematical basis of QM:   When you are given a Hamiltonian, it comes with the Hilbert space of functions of which it acts (subject to some conditions we impose, such as the Lebesgue square-integrability of functions) When you are given another Hamiltonian, there is no guarantee that it will act on the same Hilbert space Therefore, that the zeroth order Hamiltonian has eigenstates forming a complete basis for the space on which ti acts does not guarantee that these same eignestates form a complete basis for the Hilbert space on which the purturbed Hamiltonian acts.   As a simple example. I can imagine adding a small perturbing magnetic field to a Hamiltonian in which there initially is none, such that we lose the degeneracy in the eigenstates accounting for different spins of a particle. One could say that we have the same Hilbert space before and after but we lose degeneracy; or I would have been tempted to say that our initial Hilbert space did not have these extra dimensions corresponding to the spin states. So rather than the eigenstates of the basis Hamiltonian also forming a complete basis for the purtubed Hamiltonian, but having eigenvalue degeneracy which is lifted on the purtubration, I would have said that our Hilbert space has changed. After all, if we take the former view (i.e. a basis accouting for spin but with degeneracy) then it seems like for any problem we would need to know and account for EVERY possible system property that can be discerned. This doesn't seem reasonable to me....  NOTE:  1) My question was asked here previously:   although there is only a discussion and no conclusive answer given. I see that the distinction is important in relativistic QM, and the case of adding a potential onto a free particle Hamiltonian seems to be a special one. Although I am still not sure what the answer is for general purturbations.  2) Similar question here:   but it seems to completely avoid answering the question. How does DES decryption work? Is it the same as encryption or the reverse? If DES decryption is the same as encryption done in reverse order, then how can the reversed S-Box convert 4 bits into 6 bits? Embedding DLLs in a compiled executable Is it possible to embed a pre-existing DLL into a compiled C# executable (so that you only have one file to distribute)? If it is possible, how would one go about doing it?  Normally, I'm cool with just leaving the DLLs outside and having the setup program handle everything, but there have been a couple of people at work who have asked me this and I honestly don't know. What happened to Startups SE? I really enjoyed that site. I've only posted on it once, but I enjoyed seeing other people's posts and appreciated reading about entrepreneurial dilemmas. Number of ways to choose $n$ balls from $3n$ balls, distinguishable/indistinguishable. Is there evidence to suggest that Supreme Leader Snoke is Darth Plagueis? In Episode III,  explains to Anakin the power of the Dark Side to save people from death. He mentions  - the Sith who discovered this. Sidious then claims he died, which seems odd, given that kind of power.   Then in Episode VII (The Force Awakens) we find a Sith Lord who has appeared from no where. There seems to be two loose ends that fit nicely together.   My question is: Is there evidence to suggest that Supreme Leader Snoke is Darth Plagueis? I've been using the biblatex package for a while, and since updating my TeX installation (post-el capitan upgrade) I have been experiencing the following.   If the following is typed:  Such and such has been said \citep[111]{citekey}.   The following text is displayed:  Such and such has been said (Author, 2016, p. 111).   And if the following is typed:  \citet[111]{citekey} has said such and such.   The following text is displayed:  Author, (2016, p. 111) has said such and such.   The citep command is working as I want it to, but I do not want a comma placed after the author's name when using the citet command. I would like it to display as follows:  Author (2016, p. 111) has said such and such.    If I use the \citeyearpar command, it is fine, but I would rather not have to type the author's name in manually.   Here is the command placed in the preamble:  \usepackage[backend=bibtex, style=authoryear-icomp, url=true, isbn=false, natbib=true, sortcites=true, block=space, pagetracker=false, loccittracker=constrict]{biblatex}   Here is an example of an entry in my bib file (made using BibDesk):  @article{poldrack2006, Author = {Poldrack, Russell}, Journal = {Trends in Cognitive Sciences}, Number = {2}, Pages = {59--63}, Title = {{Can cognitive processes be inferred from neuroimaging data?}}, Volume = {10}, Year = {2006}}   I have also tried changing the entry to the following:  @article{poldrack2006, Author = {Russell Poldrack}, Journal = {Trends in Cognitive Sciences}, Number = {2}, Pages = {59--63}, Title = {{Can cognitive processes be inferred from neuroimaging data?}}, Volume = {10}, Year = {2006}}   The bibliography entry appears as follows, and is fine:  Poldrack, Russell (2006). “Can cognitive processes be inferred from neuroimaging data?” In: Trends in Cognitive Sciences 10.2, pp. 59–63.  Thank you for any help that can be offered. Explaining Horizontal Shifting and Scaling I always find myself wanting for a clear explanation (to a college algebra student) for the fact that horizontal transformations of graphs work in the opposite way that one might expect.  For example, $f(x+1)$ is a horizontal shift to the left (a shift toward the negative side of the $x$-axis), whereas a cursory glance would cause one to suspect that adding a positive amount should shift in the positive direction. Similarly, $f(2x)$ causes the graph to shrink horizontally, not expand.  I generally explain this by saying $x$ is getting a "head start". For example, suppose $f(x)$ has a root at $x = 5$. The graph of $f(x+1)$ is getting a unit for free, and so we only need $x = 4$ to get the same output before as before (i.e. a root). Thus, the root that used to be at $x=5$ is now at $x=4$, which is a shift to the left.  My explanation seems to help some students and mystify others. I was hoping someone else in the community had an enlightening way to explain these phenomena. Again, I emphasize that the purpose is to strengthen the student's intuition; a rigorous algebraic approach is not what I'm looking for. Currently, if you access a deleted question you will get a short message informing you that the question has been deleted. However, if you click on a direct link to a deleted answer or comment (which you can get via notifications), nothing is displayed.  This is naturally quite confusing as you will get the message about a new answer or comment via notifications, click on the link and simply see nothing.  Therefore, I propose to display some sort of message to the user, informing her/him that the answer or question has been deleted. Note that this is different from always showing a deleted answer/comment, as I specifically request it in case you access the specific URL.  One example for a comment would be:   One example for a deleted answer is:   This question is related to ,  but requests a feature to circumvent the problem altogether. Studying for an exam, a review question...  Given $G=\langle x,y|x^4=y^4=e,xyxy^{-1}=e\rangle$.    Show $|G|\leq16$.   For this, I want to consider that $x^3=x^{-1}$ and $y^3=y^{-1}$ based on our assumptions. I am a little lost as to how to put the second part, $xyxy^{-1}=e$ to show there are no more than 16 elements.   If $|G|=16$, find the center of the group and find a group that is isomorphic to $G/\langle y^2\rangle$.   I'm pretty sure the group is centerless. How to Combine two Normal Maps? I made a high poly model and then baked its textures onto a low poly model. Now I want to give another normal map's details to the model. How can this be done? Are there any rulesets or laws for Q or Q continuum? Anything, that would prevent them from doing... anything? Or at least limit them from doing something really bad?  In "" episode Q explains to  that they can do, have, get, become anything they want or appear anywhere they want, with just a snap of fingers. If that is true and if there are absolutely no borders or limits, they can't pass, then how can universe work in general? Or, how did it manage to exist so far?  I mean -- any Q could wipe out Borg, entire human population or even entire universe... with just a snap of fingers. What could prevent Q from doing that? Is there evidence that adding salt to water prior to boiling can damage a stainless steel pan? I have heard that two permutations are conjugate if they have the same cyclic structure. Is there an intuitive way to understand why this is?