A little while ago I came across this question on Yahoo! Answers and it had no replies. I have spent a while trying to figure this out as well, but I couldn't find an algebraic proof. The assumptions are $a+b=1$ and $a,b\in(0,1)$. We want to show that $a^ab^b+a^bb^a\leq 1$. Or equivalently that $a^a(1-a)^{1-a}+a^{1-a}(1-a)^a\leq 1\;\forall a\in(0,1)$.  I tried see if $a$ and $b$ are an upper bound for $a^ab^b$ and $a^bb^a$ respectively, for their sum would be $1$ by assumption. However, I was unable to do so. I managed to show the following, assuming without loss of generality that $b\leq a$ (i.e. $a\geq\frac{1}{2}$): $$\begin{aligned} a^ab^b&amp;=a^{1-b}b^b\\ &amp;=a\left(\frac{b}{a}\right)^b\\ &amp;\leq a. \end{aligned}$$ However, applying this same trick to $a^bb^a$ does not yield the necessary inequality: $$\begin{aligned} a^bb^a&amp;=a^bb^{1-b}\\ &amp;=b\left(\frac{a}{b}\right)^b\\ &amp;\geq b. \end{aligned}$$ Which means that the upperbound $a^bb^b\leq a$ is too weak. By similar means one can find the following bounds: $$b\leq a^ab^b\leq a,$$ $$b\leq a^bb^a\leq a.$$  Now define $f(a):=a^a(1-a)^{1-a}=a^ab^b$. One can show that the following holds for the derivative of $f$: $$f'(a)=f(a)\log\left(\frac{a}{1-a}\right).$$ Since $f(a)=\exp(a\log(a)+(1-a)\log(1-a))&gt;0$, we have that $f'(a)=0\iff a=\frac{1}{2}$. Since $f'(a)&gt;0$ for all $a&gt;\frac{1}{2}$ we have that $f$ has a minimum at $a=\frac{1}{2}$. Also note that $\lim_{a\to 1}f(a)=1$, hence by symmetry $\frac{1}{2}\leq f(a)\leq 1$ for all $a\in(0,1)$.  We'll define $g(a):=a^{1-a}(1-a)^a=a^bb^a$. One finds that $$g'(a)=g(a)\left(\frac{1-2a}{a(1-a)}+\log\left(\frac{1-a}{a}\right)\right).$$ Now $g'(a)=0\iff\log\left(\frac{1-a}{a}\right)=\frac{2a-1}{a(1-a)}$. It's clear that $a=\frac{1}{2}$ is a solution. Note that $\frac{d}{dx}\log\left(\frac{1-a}{a}\right)=\frac{1}{a(a-1)}&lt;0$. Since $\frac{2a-1}{a(1-a)}$ is increasing, $a=\frac{1}{2}$ is the only solution. As $\lim_{a\to 1}g(a)=0$, we have that this is a maximum. Hence $0\leq g(a)\leq\frac{1}{2}$ for all $a\in(0,1)$.  Combining these results with what we already had yields the following inequalities: $$\frac{1}{2}\leq a^ab^b\leq a,$$ $$b\leq a^bb^a\leq\frac{1}{2}.$$ This gives us $$b+\frac{1}{2}\leq a^ab^b+a^bb^a\leq a+\frac{1}{2}.$$ Which is still not all too helpful... So then I got another idea. I construct two parabolas $\phi,\psi$ both with their top at $\left(\frac{1}{2},\frac{1}{2}\right)$. One of them, $\phi$, through $(0,0)$ and $(1,0)$ and $\psi$ through $(0,1)$ and $(1,1)$. Then I'd show that $a^a(1-a)^{1-a}\leq\psi(a)$ and $a^{(1-a)}(1-a)^a\leq\phi$. Then, since $\phi(a)+\psi(a)=1\;\forall a$ the statement is proven. However, this created some difficulty that I am now stuck on. I'll first define $\phi$ and $\psi$: $$\phi:(0,1)\to\left(0,\frac{1}{2}\right]$$ $$x\mapsto-2x^2+2x$$ Then $\psi(x)=1-\phi(x)$. We have $$\begin{aligned} \phi(a)&amp;\geq a^{1-a}(1-a)^a\\ \iff-2a^2+a&amp;\geq a^{1-a}(1-a)^a\\ \iff2a&amp;\geq\left(\frac{a}{1-a}\right)^{1-a}. \end{aligned}$$ I cannot figure out how to prove this inequality though; let alone the other one. Hence now I am stuck!  I'm hoping someone can help figure this out. I realize parts of what I've done may be superfluous, but this is just everything that I have tried put together.
Given $a + b = 1$, Prove that $a^ab^b + a^bb^a \le 1$; $a$ and $b$ are positive real numbers.
Erf as term for plot of land in a town or city. Is this word only used in southern Africa? Nth max salary in Oracle I've found several posts with this same question, but the answers always diverge to some other topic or they don't work for my needs.  After a node is saved, I want my custom module to insert a row into one of my custom tables which contains the nid of the new node. When hook_node_submit is called, the nid doesn't exist yet. hook_node_insert does have the nid, but only gets called the first time the node is saved, but I potentially need to take action each time the node is saved.   My plan is to separate my functionality into its own method and implement both hooks, but I'm trying to learn to do things properly with Drupal and so I'm wondering if there is a hook that actually fits with my requirements.  Is there a hook which is called every time a node is saved and called AFTER the save is complete? Equals(=) vs. LIKE When using SQL, are there any benefits of using = in a WHERE clause instead of LIKE?  Without any special operators, LIKE and = are the same, right? Dear all I think I have a problem with the collation of makeidx.  following some suggestions found here I have got in the text something like this  \index[p]{Plutarco!demalignitate@\textit{De Malignitate Herodoti}!00855 @855A|qt}  \index[p]{Plutarco!demalignitate@\textit{De Malignitate Herodoti}!00868 @868f-869b|nn{\thefootnote}}   the first is an \index entry in the text, the second in a footnote. My problem is that even if I put the sort also on the second level of indexing it keeps keeping them separated.  I use this package information  \usepackage[splitindex]{imakeidx} \makeindex[name=n, title=Nomi di persone e luoghi]  \makeindex[name=p, title=Passi citati nel testo] \makeindex[name=t, title=Altri testi citati] \makeindex[name=pap, title=Papiri e manoscritti citati nel testo]  \newcommand{\nn}[2]{#2 \textit{n}#1} \newcommand{\qt}[1]{\textbf{#1}}   in the .idx file I get this  \indexentry{Plutarco!demalignitate@\textit{De Malignitate Herodoti}!00855 @855A|hyperindexformat{\qt}}{8}  \indexentry{Plutarco!demalignitate@\textit  {De Malignitate Herodoti}!00868 @868 b-c|hyperindexformat{\nn{33}}}{10}   can somebody advise me on what I am doing wrong here, please? I have tried to use \string after the first sort with no effect.    sure, sorry: this is the file I'm using  \documentclass[12pt,a4paper]{book} \usepackage[utf8x]{inputenx}  \usepackage{palatino}        \usepackage[LGR,T1]{fontenc} \usepackage[english,german,polutonikogreek,italian]{babel}  \usepackage[splitindex]{imakeidx} \makeindex[name=p, title=Passi citati nel testo,columns=1] \newcommand{\nn}[2]{#2 \textit{n}#1} \newcommand{\qt}[1]{\textbf{#1}} \usepackage[hidelinks]{hyperref} \begin{document}  (855A\index[p]{Plutarco!demalignitate@\textit{De Malignitate Herodoti}!00855 @855A|qt})% \footnote{DHM 868 B-C\index[p]{Plutarco!demalignitate@\textit{De Malignitate Herodoti}!00868 @868 b-c|nn{\thefootnote}}} \printindex[p] \end{document} What is a good book to study linear algebra? I'm looking for a book to learn Algebra. The programme is the following. The units marked with a $\star$ are the ones I'm most interested in (in the sense I know nothing about) and those with a $\circ$ are those which I'm mildly comfortable with. The ones that aren't marked shouldn't be of importance. Any important topic inside a unite will be boldfaced.  U1: Vector Algebra. Points in the $n$-dimensional space. Vectors. Scalar product. Norm. Lines and planes. Vectorial product.  $\circ$ U2: Vector Spaces. Definition. Subspaces. Linear independence. Linear combination. Generating systems. Basis. Dimesion. Sum and intersection of subspaces. Direct sum. Spaces with inner products.  $\circ$  U3: Matrices and determinants. Matrix Spaces. Sum and product of matrices. Linear ecuations. Gauss-Jordan elimination. Range. Roché Frobenius Theorem. Determinants. Properties. Determinant of a product. Determinants and inverses.  $\star$ U4: Linear transformations. Definition. Nucleus and image. Monomorphisms, epimorphisms and isomorphisms. Composition of linear transformations. Inverse linear tranforms.  U5: Complex numbers and polynomials. Complex numbers. Operations. Binomial and trigonometric form. De Möivre's Theorem.  Solving equations. Polynomials. Degree. Operations. Roots. Remainder theorem. Factorial decomposition. FTA. Lagrange interpolation.  $\star$ U6: Linear transformations and matrices. Matrix of a linear transformation. Matrix of the composition. Matrix of the inverse. Base changes.   $\star$ U7: Eigen values and eigen vectors Eigen values and eigen vectors. Characteristc polynomial. Aplications. Invariant subspaces. Diagonalization.   To let you know, I own a copy of Apostol's Calculus $\mathrm I $ which has some of those topics, precisely:   Linear Spaces Linear Transformations and Matrices.   I also have a copy of Apostol's second book of Calc $\mathrm II$which continues with   Determinants Eigenvalues and eigenvectors Eigenvalues of operators in Euclidean spaces.   I was reccommended Linear Algebra by Armando Rojo and have Linear Algebra by , which seems quite a good text.   What do you reccomend? How can I display additional boot and shutdown information on the Windows 7 welcome screen? There is a small tweak, I believe it is a registry key, that allows to display additional information on the Welcome and Shutting down screens of Windows 7 (and most likely Vista, too). I have activated this tweak on one of my systems; unfortunately I forgot how I did it, and I can't seem to find the website that originally gave me that information.  Usually, the Windows 7 welcome screen will just display "Welcome" when logging in. With the tweak activated, my Welcome screen gives status information such as "Loading user settings" or "Preparing desktop". When shutting down, the default screen simply says "Shutting down". With the tweak activated, it gives additional status information such as "Stopping Windows services". 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Login screen "Waiting for User Profile Service"   Login screen "Preparing desktop"   Logout screen "Stopping Windows services": Book about a teenager programmed to be an assassin and trying to get back at the people responsible I read this book a few years back (around 5?). It was part of a multibook series but I have only read one book.  It's about this teenager who I think had been programmed to be an assassin but then somehow retained his conscience and was out to get the people responsible for that.  I briefly remember a girl, and this one part where he realizes he can ride a horse and then there's a chase too. I think the guy's name was Jim or Jimmy, but not quite. I'm also quite sure he was either part computer, or had a chip or something in his brain which had been programmed to give him a host of mental and physical abilities which he retained and used as well as the series progressed. According to the , when a Shinigami dies, their remaining lifespan is given to the human they saved.     Shinigami who die are reduced to dust, and their remaining lifespan is given to the human they saved.   Since Rem killed L and Watari, she gained their remaining lifespans. Additionally she probably still had some lifespan left. On top of that Misa was already given a lot of lifespan from Gelus (from which she lost 75% again). So does this mean all of this lifespan is transferred to Misa Amane?!  Yet according to her , she dies in 2011 at the age of 26, only seven years after Rem died.  Does this mean that humans are also capable of reducing their own lifespans? Or would the author have made a mistake by letting Misa die as early as she has? I know she lost 75% of her remaining lifespan through the two eye deals, but you'd think that the cumulation of L, Watari and Rem's remaining lifespans would be greater than just one year, wouldn't you? 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Is there a way to set this folder as the default so that Windows Explorer always opens to this folder? problem with surfaces on tkz-fct i am using this wonderful tool that tkz-fct is for creating nice documents but I have a weird result from recently. This is my document :  \documentclass[10pt,a4paper]{article} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage{calc} \usepackage{color} \usepackage{pifont}                     %pour utiliser la police Ding \usepackage[left=1.5cm,right=1.5cm,top=2cm,bottom=2cm]{geometry} %marges du document \usepackage[french]{babel} \usepackage[usenames,dvipsnames]{xcolor} \usepackage[L]{thmbox} \usepackage{eurosym} \usepackage{enumitem}               %pour modifer itemize et enumerate \usepackage{amsmath,amsfonts,amssymb} \usepackage{mathrsfs}               %pour utiliser mathscr \usepackage{array,multirow,makecell} \usepackage{fancyhdr}               %package pour modifier les en-têtes \usepackage{fancybox}               %pour boîtes spéciales \usepackage{tikz,tkz-tab,tkz-fct} \usepackage{tikz-dependency}        %permet de créer des schemas \usepackage{sectsty}                %permet de modifier les styles des sections (voir \sectionfont l.21) \usepackage{fourier-orns} \usepackage{shadethm}               %pour theoremes en gris \setcellgapes{1pt} \makegapedcells \newcolumntype{R}[1]{&gt;{\raggedleft\arraybackslash }b{#1}}   %pour redéfinir les colonnes dans tabular \newcolumntype{L}[1]{&gt;{\raggedright\arraybackslash }b{#1}} \newcolumntype{C}[1]{&gt;{\centering\arraybackslash }b{#1}} \renewcommand{\thesection}{\Roman{section}} %permet de changer la numérotation en chiffres romains des sections \newshadetheorem{defi}{Définition}  %crée un nouveau type de théorème \newshadetheorem{thm}{Théorème} \newshadetheorem{ppte}{Propriété} \sectionfont{\sectionrule{3ex}{3pt}{-1.5ex}{1pt}} %mettre un trait de séparation dans les sections (package sectsty) \pagestyle{fancy} %style en-têtes \lhead{} %en-tête gauche \chead{Lois à densité} %en-tête centrale \rhead{T\up{ale}} %en-tête droite \begin{document} \begin{tikzpicture}      \tkzInit[xmin=-4,xmax=4,ymax=1]      \tkzDrawX[label={},noticks]      %\tkzDrawY      \tkzFct[domain=-2:2]{exp(-x**2)}      \tkzDrawArea[color=blue, domain = -0.5:1]      \tkzText(-1,0.75){$\mathscr{C}_{f}$} \end{tikzpicture} \end{document}   But from now, it doesn't produce the result I would like as the surface start from the point (-0.5,f(-0.5)) and not as an integral from -0.5 to 1. This is a picture of the result :   Which is obviously not what I would like and this extends now to all my "old" docs which were "normal" before. I don't think I have done anything special except updating my Tex distribution ... If someone has an idea ?? I was presented with the following problem:      Prove that there exist solutions to $x^2+y^2=z^n$ for all $n$, with $x,y,z, n \in \mathbb{N}$   I showed that by taking any Pythagorean triple $x^2+y^2=z^2$ and multiplying by $z^{2(n-1)}$ we get $(z^{n-1}x)^2+(z^{n-1}y)^2=(z^2)^n$, which allows me to generate solutions easily for any value of $n$. I managed to find several similar questions on this site, such as  concerning the specific case $n=3$. I notice that all of these questions take a similar approach and start with a Pythagorean triple and use it to generate general solutions.  Is there a way to prove the statement (or better yet provide solutions to the equation) without first relying on Pythagoras? I have installed Ubuntu on my laptop and now I want to remove it and the Grub Loader. How I can do it without the Windows installation disk (I cannot use the Recovery Console).