If $A.M$ and $G.M$ are Arithmetic Mean And Geometric Mean respectively then prove that $A.M \ge G.M$.  My Attempt :  Let $a$ and $b$ are any two real positive numbers.  Then: $$A.M=\frac{a+b}{2}$$  $$G.M =\sqrt{ab} $$  Now how can we show that $A.M.\ge G.M.$ ? Please help me..
The arithmetic - geometric mean inequality states that $$\frac{x_1+ \ldots + x_n}{n} \geq \sqrt[n]{x_1 \cdots x_n}$$ I'm looking for some original proofs of this inequality. I can find the usual proofs on the internet but I was wondering if someone knew a proof that is unexpected in some way. e.g. can you link the theorem to some famous theorem, can you find a non-trivial geometric proof (I can find some of those), proofs that use theory that doesn't link to this inequality at first sight (e.g. differential equations …)?  Induction, backward induction, use of Jensen inequality, swapping terms, use of Lagrange multiplier, proof using thermodynamics (yeah, I know, it's rather some physical argument that this theorem might be true, not really a proof), convexity, … are some of the proofs I know.
INT vs Unique-Identifier for ID field in database Floating point comparison "Speed" of Gravity and Speed of Light What is the best way to clean the sensor on a interchangeable lens camera (mirrorless or digital SLR)? I've got a couple of spots of dust on the sensor of my Canon 20D. A rocket blower isn't shifting them, and I'm ridiculously scared of some of the more direct methods for cleaning sensors — what methods/systems have you tried, and how successful have they been? Creating or referencing variables dynamically in Sass Checking out  and  (and some papers) was a bit difficult for understanding Group Action. The closest to understanding was this part:    I understand the definition of homomorphism (a structure-preserving map between two algebraic structures of the same type). And from what I understand the symmetric group is a group of functions (bijective functions) mapping elements of a set $X$ onto itself.  However I get confused at this line:     ...an action of $G$ on $X$ may be formally defined as a group homomorphism $\phi$ from $G$ to the symmetric group of $X$.   That seems to be the definition of group action, but I am having trouble understanding how to apply that.  What that says to me is it is a homomorphism (function) $f \colon X \rightarrow S$ where $S$ is a set of bijective functions on $X$. That itself seems weird, I thought a group action would result in a number or something related to $\{1, 2, 3\}$ (defined next).  Continuing, if we have a set $X = \{1, 2, 3\}$ and a group $(*,\{4, 5, 6\})$, then the symmetric group of $X$ would be something like $S = \{1 \rightarrow 2, 2 \rightarrow 3, \dots \}$, and $\phi$ would be something like $f : \{4, 5, 6\} \rightarrow S$. That doesn't make sense to me and I feel like I'm interpreting something wrong. I don't see how mapping to a set of functions $S$ will result in an action on the set. delete vs delete[] operators in C++ If a lens focuses all incoming light to a point, how do we get 2D images? How do lenses produce 2-dimensional images, if a lens bends all incoming rays of light to intersect at the focal point? Shouldn't this produce a single dot of light on a screen placed at the focal length?  This is basically the standard diagram that always shows up in textbooks:    I know this doesn't happen in real life--I used to use telescopes pretty frequently for work. The most in-focus image would be the one with the smallest diameters for the stars, which, of course, we think of point sources at infinity. But--the light from all stars (despite being at infinity) is not all focused at a single point. Instead, each star's light is focused at its own $(x,y)$ point on a 2D image.  I can't reconcile the theory as I understand it with any of my real-life experience with optics. What am I missing? Change paper size in mid-document My document should be letterpaper and my attachments should be a4paper. I had high hopes for the \newgeometry command from the geometry package, but the documentation says      \newgeometry is almost similar to   \geometry except that \newgeometry   disables all the options specified in   the preamble and skips the   papersize-related options: landscape,   portrait and paper size options (such   as papersize, paper=a4paper and so   forth).   Any help is appreciated! I'm hoping someone can help! I recently downloaded a torrent directly onto an external hard drive, (using a 2009 MacBook Pro and 1TB Seagate Expansion portable Drive) somehow during the download the drive was accidentally ejected, since this incident when I plug the drive back in it’s having problems reading.  The drive works fine for about 10 minutes, I can read/copy/paste and access files but then starts to wreak havoc; apps hangs, it hangs the Finder, etc.  I’ve tried repairing it in Disk Utility, which reads and repairs this disk but problems inevitably continue as described above.  I’ve deleted the Torrent file also. All my files on this external drive so re-formatting is not an option until I can retrieve them. How to remove illegal characters from path and filenames? I need a robust and simple way to remove illegal path and file characters from a simple string. I've used the below code but it doesn't seem to do anything, what am I missing?  using System; using System.IO;  namespace ConsoleApplication1 {     class Program     {         static void Main(string[] args)         {             string illegal = "\"M&lt;&gt;\"\\a/ry/ h**ad:&gt;&gt; a\\/:*?\"&lt;&gt;| li*tt|le|| la\"mb.?";              illegal = illegal.Trim(Path.GetInvalidFileNameChars());             illegal = illegal.Trim(Path.GetInvalidPathChars());              Console.WriteLine(illegal);             Console.ReadLine();         }     } } Add layer to a QGIS group using Python I am trying to add new layers into a specific group which I can get to work if I know the index, but how can I get Python to find the index of a group so it can load the new layer into the group wherever it is in the TOC?  qgis.utils.iface.legendInterface().moveLayer( layer, index ) How to programmatically create a WPF window in a WinForm application Two friends who have unpredictable lunch hours agree to meet for lunch at their favorite restaurant whenever possible. Neither wishes to eat alone and each dislikes waiting for the other, so they agree that each will arrive at a random time between noon and 1 pm, and each will wait for the other for 15 minutes or until 1:00. What is the probability that the friends will meet for lunch on a given day? How to import an FBX file into Blender I am trying to import an FBX file from Unity into Blender. Does anybody know how to do that? javascript document.getElementsByClassName compatibility with IE Is it possible to programmatically stop and start ArcGIS Server map services using Python? Why does default auto.arima stop at (5,2,5)? Show that $S = \sqrt{S^2}$ is a biased estimator of $\sigma$ given a random sample from a normal distribution ... Suppose $Y_1, \ldots, Y_n$ is a random sample from a normal distribution with mean $\mu$ and variance $\sigma^2$. Let $S^2$ be the sample variance, which is unbiased for $\sigma^2$.  GOAL: Show that $S = \sqrt{S^2}$ is a biased estimator of $\sigma$.  I realize that $(n-1)S^2/ \sigma ^2$ is chi-square with $(n-1)$ degrees of freedom.  I am given that $$E(S) = E\left[\frac{\sigma}{\sqrt{n-1}} \left[\frac{(n-1)S^2}{\sigma^2}\right]^{1/2}\right] = \frac{\sigma}{\sqrt{n-1}} \int_0^\infty v^{1/2} \frac{1}{\Gamma[(n-1)/2] 2^{(n-1)/2}} v^{(n-1)/2} e^{-v/2} \, dv$$  Can anyone help me understand why they have $v^{1/2}$ and $v^{(n-1)/2}$ (combining them $v^{n/2}$ as I think it should be $v^{1/2}$ and $v^{\frac{(n-1)}{2}-1}$ (combining both terms give $v^{(n-2)/2}$ since the pdf for chi-square distribution is $$ f(v) = \frac{v^{(x/2)-1} e^{-v/2}}{2^{x/2}\Gamma(x/2)},$$ where $x$ is degree of freedom?  Could anyone please tell me which one is the correct answer with explanation?  Thank you I only have access to the server via a terminal and I can't use graphical tools such as GParted! I want to create a new partition from a part of the root (about 768mb) for swap.  # df -h Filesystem      Size  Used Avail Use% Mounted on /dev/vda         20G  1.3G   18G   7% / udev             10M     0   10M   0% /dev tmpfs           199M  4.9M  194M   3% /run tmpfs           100M   12K  100M   1% /run/user tmpfs           5.0M  4.0K  5.0M   1% /run/lock