Let $a_1,\dots,a_d,b_1,\dots,b_d$ be positive real numbers. Then $$ \prod_{i=1}^d a_i+\prod_{i=1}^d b_i \le \prod_{i=1}^d (a_i^d+b_i^d)^{1/d}$$ and equality holds if and only if $\frac{a_i}{a_j}=\frac{b_i}{b_j}$.     Is there a nice proof for this fact? Is there some geometric interpretation?   I know a proof based on an application of Holder's inequality which seems rather miraculous to me. The idea is to define $f_i:\{0,1\} \to \mathbb{R}$, by $f_i(0)=a_i,f_i(1)=b_i$, and to endow the domain $\{0,1\}$ with the counting measure. The generalized Holder's inequality then states that  $$\prod_{i=1}^d a_i+\prod_{i=1}^d b_i\le\|  \prod_{i=1}^d f_i\|_1 \leq \prod_{i=1}^d \|f_i\|_d=\prod_{i=1}^d (a_i^d+b_i^d)^{1/d}$$  and equality holds if and only if $\frac{f_i}{f_j}$ is constant, i.e. $\frac{a_i}{a_j}=\frac{b_i}{b_j}$. Somehow, I feel that it's weird to use an "abstract" measure-theoretic inequality, applied to such a degenerate case (finite measure space with the counting  measure).     Are there other, "more elementary" proofs?   Even the case where $a_i=c$ are constant does not seem trivial to me: i.e. why  $$ c^d+\prod_{i=1}^d b_i \le \prod_{i=1}^d (c^d+b_i^d)^{1/d}$$ and equality holds if and only if $b_1=b_2=\dots=b_d$.  I tried using the generalized Young's inequality for multiple products, which states that $$ \prod_{i=1}^d a_i \le \frac{\sum_i {a_i^d}}{d}.$$ (This is in fact the AM-GM inequality in disguise.)  This implies  $$ \prod_{i=1}^d a_i+\prod_{i=1}^d b_i \le \frac{\sum_i {a_i^d}+{b_i^d}}{d} \ge \prod_{i=1}^d (a_i^d+b_i^d)^{1/d}$$  but the direction of the last inequality (also by AM-GM) is opposite to the one we would like to have.  Of course, we can always mimic the (inductive) proof of the generalized Holder inequality, but this does not seem enlightening to me.
consider positive numbers $a_1,a_2,a_3,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$. does the following in-equality holds and if it does then how to prove it  $\left[(a_1+b_1)(a_2+b_2)\cdots(a_n+b_n)\right]^{1/n}\ge \left(a_1a_2\cdots a_n\right)^{1/n}+\left(b_1b_2\cdots b_n\right)^{1/n}$
Add layer to a QGIS group using Python I have a question about the following passage on pg. 89 of Zee's QFT in a nutshell:     At first sight, what Dirac wanted does not make sense. The equation is supposed to have the form "some linear combination of $\partial_\mu$ acting on some field $\psi$ is equal to some constant times the field." Denote the linear combination by $c^\mu\partial_\mu$. If the $c^\mu$'s are four ordinary numbers, then the four-vector $c^\mu$ defines some direction and the equation cannot be Lorentz invariant.   The simplest equation I can construct that he is referring to is  $$(c^{\mu}\partial_{\mu}-m)\phi(x)=0, $$  for $m&gt;0$. Taking $\phi(x)\rightarrow\phi(\Lambda^{-1}x)$, and $c^{\mu}\rightarrow\Lambda^{\mu}_{\ \ \nu}c^{\nu}$. Then, the left term transforms like   \begin{align*} \Lambda^{\mu}_{\ \ \nu}c^{\nu}(\Lambda^{-1})^{\sigma}_{\ \ \mu}\partial_{\sigma}\phi(\Lambda^{-1}x) &amp;= \delta^{\sigma}_{\ \ \nu}c^{\nu}\partial_{\sigma}\phi(\Lambda^{-1}x)\\ &amp;=c^{\sigma}\partial_{\sigma}\phi(\Lambda^{-1}x). \end{align*}  Then, we have that   $$ (c^{\sigma}\partial_{\sigma}-m)\phi(\Lambda^{-1}x)=0.$$  But doesn't this show that the equation is Lorentz-invariant? Can a willing lycanthrope be True Polymorphed? Can a lycanthrope or other shapeshifter be willingly affected by the spell True Polymorph? I know that it can be affected by regular Polymorph, but the text for the enhanced version seems to differ. After reading  It seems that there is no perfect solution for getting rid of the stretch at the poles of UV spheres.  According to , icospheres can be used for stretch-less UV mapping:     All faces have the same area, which may be useful for certain types of UV mapping containing non-organic textures. An example that comes to mind is an isocahedral die or billiard balls where stretch must be minimized near the point where the number is printed onto the ball.   Incidentally, a lowish-poly billiard ball is exactly what I'm attempting to create.  However, my attempts at neatly unwrapping an icosphere have mostly failed (smart UV unwrap works perfectly for using baked textures, however I was hoping for something more like the example in the 6th post of , with a nice clean rectangular map).  How can an icosphere (I assumed the term "geosphere" in the thread was referring to a (geodesic) sphere made of triangles, like an icosphere) be cleanly unwrapped? Or if it's not possible, is there another way to do this which I'm missing?    I'm planning on using the resulting sphere in the BGE. The script works fine if I run it with sh filename, but it ignores all the --exclude parameters if its run by cron:  #!/bin/sh TIMESTAMP=$(date +%u)  WEEK=`date +"%V"` if [ $(($WEEK%2)) -eq 0 ]; then     echo "even";     BACKUP_DIR="/mnt/backup/2_$TIMESTAMP"     mkdir -p "$BACKUP_DIR/www"  else     echo "odd";     BACKUP_DIR="/mnt/backup/1_$TIMESTAMP"     mkdir -p "$BACKUP_DIR/www"   fi  if [ $(date +%u) -gt 6 ]; then     zip -r -y $BACKUP_DIR/www/web.zip /var/www/ --exclude='var/www/\.opcache/*' --exclude='var/www/utils/phpThumb/cache/*' else     zip -r -y $BACKUP_DIR/www/web.zip /var/www/ --exclude='var/www/mycompany/files/*' --exclude='var/www/\.opcache/*' --exclude='var/www/mycompany/szamlazz/xml_szamlak/*' --exclude='var/www/mycompany/files_admin/szamlazz/*' --exclude='var/www/mycompany/files_admin/chatFiles/*' --exclude='var/www/mycompany/utils/phpThumb/cache/*' --exclude='*/backup-guard/*' --exclude='var/www/mycompany/mailer/*' --exclude='var/www/mycompany/attachments/*' fi  exit 0   The above script should create a larger backup on sunday, and a partial on weekdays, but it just zips everything. Explain what is a transistor in plain English! I am creating a type of adventure map. Once you enter the map, you step on some pressure plates that are supposed to seamlessly close the wall behind you. For some reason this command does not work when in a command block or just in text. Here is the command:  /fill -17 4 -109 -18 5 109 minecraft:light_gray_concrete   The output it gives me is "That position is not loaded". I am on a singleplayer superflat world in 1.14.3. Any help would be very appreciated! I've a set of regions within a city. I'd like to split each region into five subregions and then get the centroid of each one of them.  I know how to make a vector grid, but the problem I have is that the generated points are not taking into account the regions, they're only related to the general map.  I have this:    And I want to split each region in five equal size areas. Something like this:    I'd like to build a visualization similar to this one: . In order to accomplish that, I need to split each region, because the original ones are pretty big. Hearing a sound backwards because of Doppler effect Centering vertically in table cell I'm new with LaTeX and I'm trying to make a table with the elements centered vertically and horizontally. I've searched for this question, it seems like most people were trying to do things a little more complicated than this, resulting in complicated answers I couldn't understand. The table below is already centered horizontally, but the elements are aligned to the top of the cells instead of the center. How do I fix this?      \begin{tabular}{*{5}{|c}|}         \hline         $p$ &amp; $q$ &amp; $p \land q$ &amp; $p \lor q$ &amp; $p \to q$\\         \hline         T &amp; T &amp; T &amp; T &amp; T\\         \hline         T &amp; F &amp; F &amp; T &amp; F\\         \hline         F &amp; T &amp; F &amp; T &amp; T\\         \hline         F &amp; F &amp; F &amp; F &amp; T\\         \hline     \end{tabular} Suppose that $f(z)$ is analytic function in the disk $|z|\le R$ and $A(r)=\max_{|z|=r}Re(z)$ then prove that $r&lt;R$  $$\max_{|z|=r}\left|\frac{f^n(z)}{n!}\right|\le \frac{2^{n+2}R}{(R-r)^{n+1}}\{A(r)+|f(0)|\}$$  MY attempt   Using Cauchy generalized integral formula   $f^{(n)}(z_0)=\frac{n!}{2\pi i}\int _c\frac{f(z)dz}{(z-z_0)^{n+1}}$  then $\max_{|z|=r}\left|\frac{f^n(z)}{n!}\right|=\left|\max_{|z|=r} \frac{1}{2\pi i}\int _c\frac{f(z)dz}{(z-z_0)^{n+1}}\right|$  and i think we can solve by using Borel caratheodory theorem  Let $f(z)$ be holomorphic function on $|z|\le R$ and let $M(r)=\max_{|z|=r}|f(z)|$ and $A(r)=\max_{|z|=r}Re f(z)$ then  for $0&lt;r&lt;R$  $$ M(r)\le \frac{2r}{R-r}A(R)+\frac{R+r}{R-r}|f(0)|$$  from here how to processed ? Any time I am inserting new point feature in QGIS, the Feature Attribute dialog box is popping up. Is there a way to prevent it? (I am willing to insert many points at once and this pop up box slows down the process). Can a satellite remain directly over a city? Is it possible to put an artificial satellite into an orbit in such a way that it will always remain direct over a city (I mean at any specific place)? Find the smallest value of $c$ such that $(1-x)\ln(1-x)+(1+x)\ln(1+x)\leq c x^2$ holds for $0&lt;x&lt;1$.   I saw the curve and realized this is true for $c=2$. How can I prove it? What is the smallest $c$ that still makes the inequality valid?  I think we should use the Taylor expansions of the $\ln$s. Can you enter the Schengen area with a UK visa? Which Stack Exchange sites use MathJax? As the title says, I could use an exhaustive list of all Stack Exchange sites that use MathJax.  If the list included any site-specific customizations made to the MathJax configuration on each site, that would be even better. What is the difference between "citizen" and "denizen" Citizen: 1. A legally recognized subject ornational of a state or commonwealth, either native or naturalized. 2. An inhabitant of a particular town or city.  Denizen: 1. An inhabitant or occupant of a particular place.  Same thing? Prove the equality: $\det\left[\begin{smallmatrix} -2a &a+b &a+c \\ b+a&-2b &b+c \\ c+a&c+b &-2c \end{smallmatrix}\right] = 4(a+b)(b+c)(c+a)$ i have to prove:   $$\begin{vmatrix} -2a &amp;a+b  &amp;a+c \\   b+a&amp;-2b  &amp;b+c \\   c+a&amp;c+b  &amp;-2c  \end{vmatrix} = 4(a+b)(b+c)(c+a)$$   I have tried many calculations between the rows and columns of the determinant to get to the answer i want to solve the exercise however none of them gave me something right. Can someone help? Several times now I have inadvertently lost the .tex file that generated a PDF file with pdflatex. Given what I know of the internal structure of PDF files, it should be possible to embed a text file containing the original LaTeX file into the PDF file so that it can be extracted later.   I am aware of no technology for doing this, but perhaps you are. Does anybody have something like this? Python Script returns unintended "None" after execution of a function