Prove $\displaystyle\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{d+3}+\frac{d}{a+3}\le 1$. Given $a,b,c,d\ge 0$ and $a^2+b^2+c^2+d^2=4$ show the following holds : $$\displaystyle\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{d+3}+\frac{d}{a+3}\le 1$$ Now I tried to fully expand but that becomes too ugly. I want a non-expand proof of this fact. For some motivation I solved a three variable version which I assumed to be true and it came out to be true! That one is : $a,b,c\ge 0$ and $a^2+b^2+c^2=3$ show that : $\displaystyle \frac{a}{b+2}+\frac{b}{c+2}+\frac{c}{a+2}\le 1$. The second one is easier and I DO NOT NEED a solution of the second one. Please help me out on the first one because it is similar but not solvable by a similar method.
How prove this inequality $\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{d+3}+\frac{d}{a+3}\le 1$ Question:  let $a,b,c,d\ge 0$,such $$a^2+b^2+c^2+d^2=4$$ show that $$\dfrac{a}{b+3}+\dfrac{b}{c+3}+\dfrac{c}{d+3}+\dfrac{d}{a+3}\le 1$$  My try: By Cauchy-Schwarz inequality,we have $$\sum_{cyc}\dfrac{a}{b+3}\le\sqrt{(\sum_{cyc}a^2)\left(\sum_{cyc}\dfrac{1}{(b+3)^2}\right)}$$ then we have only prove this $$\sum_{cyc}\dfrac{1}{(a+3)^2}\le \dfrac{1}{4}?$$ This is  not true,in fact, we have $$\sum_{cyc}\dfrac{1}{(a+3)^2}\ge \dfrac{1}{4}?$$ because we have $$\dfrac{1}{(a+3)^2}\ge\dfrac{5-a^2}{64}$$ this is true because $$\Longleftrightarrow \dfrac{(a-1)^2(a^2+8a+19)}{64(a+3)^2}\ge 0$$ so $$\sum_{cyc}\dfrac{1}{(a+3)^2}\ge\sum_{cyc}\dfrac{5-a^2}{64}=\dfrac{1}{4}$$ can see:  This methods is from: can  see:  By the way I have see this three variable inequality  let $a,b,c$ be non-negative numbers such that $$a^2+b^2+c^2=3$$  show that $$\dfrac{a}{b+2}+\dfrac{b}{c+2}+\dfrac{c}{a+2}\le 1$$  proof: By expanding,the inequality becomes $$ab^2+bc^2+ca^2\le abc+2$$ without loss of generality,assume that  $$\min(a,b,c)\le b\max(a,b,c)$$ then \begin{align*} 2-ab^2-bc^2-ca^2+abc&amp;=2-ab^2-b(3-a^2-b^2)-ca^2+abc\\ &amp;=(b^3-3b+2)-a(b^2-ab+ca-bc)\\ &amp;=(b-1)^2(b+2)-a(b-a)(b-c)\ge 0 \end{align*} Equality occurs for $(a,b,c)=(1,1,1)$ and also for $(a,b,c)=(0,1,\sqrt{2})$ or any cyclic permutation.  so my Four-inequality  variable inequality,can use this methods $$\dfrac{a}{b+3}+\dfrac{b}{c+3}+\dfrac{c}{d+3}+\dfrac{d}{a+3}\le 1$$ $$\Longleftrightarrow a^2cd+3a^2c+3a^2d+9a^2+ab^2d+3ab^2+abc^2-abcd+3ac^2+9ac+3b^2d+9b^2+3bc^2+bcd^2+3bd^2+9bd+9c^2+3cd^2+9d^2-81\le 0$$ $$\Longleftrightarrow a^2cd+3a^2c+3a^2d+ab^2d+3ab^2+abc^2+3ac^2+9ac+3b^2d+3bc^2+bcd^2+3bd^2+9bd+3cd^2\le 45$$ $$\Longleftrightarrow a^2(cd+3c+3d)+ab^2d+3ab^2+abc^2+3ac^2+9ac+3b^2d+3bc^2+bcd^2+3bd^2+9bd+3cd^2\le 45$$ $$\Longleftrightarrow (4-b^2-c^2-d^2)(cd+3c+3d)+ab^2d+3ab^2+abc^2+3ac^2+9ac+3b^2d+3bc^2+bcd^2+3bd^2+9bd+3cd^2\le 45$$  Then  I can't  Thank you  very  much
Convert .sdf database to .mdf database Why are most of the main characters White/English in Game Of Thrones? Automorphism group of the quaternion group Let $Q_8$ be the quaternion group. How do we determine the automorphism group $Aut(Q_8)$ of $Q_8$ algebraically? I searched for this problem on internet. I found some geometric proofs that $Aut(Q_8)$ is isomorphic to the rotation group of a cube, hence it is isomorphic to the symmetric group $S_4$. I would like to know an algebraic proof that $Aut(Q_8)$ is isomorphic to $S_4$. How do I find my follower if and when they leave me? I had parked my follower at an inn with a bunch of valuable items that I didn't want to carry. I guess I have been out a bit too long as this message appears:      Your follower tires of waiting and leaves.    There have been other situations where my follower has mysteriously disappeared as well.  How can I find my follower when they get up and leave or disappear into thin air? Which tools and technologies are used to build the Stack Exchange Network? What tools and technologies are used to build the Stack Exchange Network?  See also: How to install KDE along Unity? I have just installed the latest Ubuntu 14.04 and I would like to install KDE along side my Gnome/Unity that I have now. How can I do this?  Thanks! $a,b,c$ are distinct real numbers that $$(a^2)(1-b+c)+(b^2)(1-c+a)+(c^2)(1-a+b)=ab+bc+ca.$$ Prove $$\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}=1.$$ I have tried two different approaches to this problem but am stuck.  First I started by expanding the bottom of each fraction. $\frac{1}{(a^2)-2ab+(b^2)}+\frac{1}{(b^2)-2bc+(c^2)}+\frac{1}{(c^2)-2ac+(a^2)}=1$  I also tried to expand $(a^2)(1-b+c)+(b^2)(1-c-a)+(c^2)(1-a+b)=ab+bc+ca$   $(a^2)-(a^2)b+(a^2)c+(b^2)-(b^2)c-(b^2)a+(c^2)-(c^2)a+(c^2)b=ab+bc+ca$ $(a^2)-(a^2)b+(a^2)c+(b^2)-(b^2)c-(b^2)a+(c^2)-(c^2)a+(c^2)b-ab-bc-ca=0$ Then I added one to both sides $(a^2)-(a^2)b+(a^2)c+(b^2)-(b^2)c-(b^2)a+(c^2)-(c^2)a+(c^2)b-ab-bc-ca+1=0+1$ Help will be appreciated. $4$ letter words taken from the letters of CONCENTRATIONS How many words with or without meaning can be made from the letters of the word CONCENTRATIONS by taking $4$ letters at a time?  There are  $14$ letters total present in the word. $4$ letters can be picked in $14\choose 4$ ways. $4$ letters can be arranged in $4!$ ways. So ideally answer has to be ${14\choose 4} * 4!$. But I know this answer is wrong because some letters are repeating. How do I solve this then? I appreciate any help. I am a Indian citizen in the United States on a F-1 visa.  For the summer break, I am traveling (round-trip) from Chicago to New Delhi via Dusseldorf-Frankfurt by Lufthansa airlines. My onward Journey would be on July - 8 and return Journey on Aug - 11.  My layover will be of 1 hour 10 mins in Frankfurt and 3 hours in Dusseldorf.  I would like to know if I require Category A or Category B Visa. If not, what type of visa do I require for the Schengen visa for my transit. How I can apply for the same?  I tried to find slots in German consulate but all slots are full before my departure date. If I don't get the Schengen visa, I cannot travel and will loose all the fare (since my ticket is a non-refundable ticket). Is there any other way I can tackle this situation. Binomial confidence interval estimation - why is it not symmetric? I have used the following r code to estimate the confidence intervals of a binomial proportion because I understand that that substitutes for a "power calculation" when designing receiver operating characteristic curve designs looking at detection of diseases in a population.  n is 150, and the disease, we believe, is 25% prevalent in the population. I have calculated the values for 75% sensitivity and 90% specificity (because that's what people seem to do).      binom.test(c(29,9), p=0.75, alternative=c("t"), conf.level=0.95)      binom.test(c(100, 12), p=0.90, alternative=c("t"), conf.level=0.95)   I have also visited this site:    Which is a java page which calculates binomial confidence intervals, and it gives the same answer.  Anyway, after that lengthy set-up, I want to ask why the confidence intervals are not symmetric, e.g. sensitivity is      95 percent confidence interval:    0.5975876 0.8855583      sample estimate probability: 0.7631579    Sorry if this is a basic question, but everywhere I look seems to suggest that they will be symmetric, and a colleague of mine seems to think they will be too. example of compact operator Let $E=l^p=\left\{x=\left(x_n\right): x_n\in\mathbb{R}, \sum_{1}^{\infty}\left|x_n\right|^p&lt;\infty\right\}$ with $1\le p&lt;\infty$, $\left\|x\right\|_E=\left(\sum_{1}^{\infty}\left|x_n\right|^p\right)^{1/p}.$ Let $\left(\lambda_n\right)$ is a bounded sequence in $\mathbb{R}$ and consider the operator $T$ defined by $$T(x)=\left(\lambda_1x_x,\ldots,\lambda_nx_n,\ldots\right)$$ where $x\in l^p$. Prove that $T$ is a compact operator from $E$ to $E$ iff $\lambda_n\rightarrow 0$.  Can anyone give me some hints to solve this problem? Thank you! Cruise ship in the Schengen area: Do the days spent at sea count toward the limit for short visits? I am planning on spending 90 days in Schengen zone countries.  I will be taking a cruise between Lisbon and Barcelona.  Do the days at sea count against the 90 days allowed for staying in the Schengen zone? Will running a large query on a secondary database in an availability group affect transaction performance in the primary database? I need to provide real-time, or almost real-time, data for SSRS and Tableau reporting. I don't want the production OLTP system to be negatively impacted by long running queries. Will running a large query on a secondary database in an availability group affect transaction performance in the primary database? Unable to lock the administration directory (/var/lib/dpkg/) is another process using it? I want to do this using the Math.Round function Prove $GL_n(\Bbb{R})/SL_n(\Bbb{R}) \cong \Bbb{R}^\times$ $$GL_n(\Bbb{R})/SL_n(\Bbb{R}) \cong \Bbb{R}^\times$$  This is trivial to prove with the first isomorphism theorem - by using the determinant as a endomorphism, then as $SL_n(\Bbb{R})$ is $1$ under the determinant, it is the kernel, and by FIT, the above is proved.  I was wondering how to prove this without the isomorphism theorem ? How do I convert materials from Blender Internal to Cycles? I know this may be a tough or confusing question, but I have numerous existing materials I created and reuse for a project, which were all prepared for Blender Internal.  Now, I want to start the transition to Cycles, but I've hit a roadblock. How can I transition my materials without completely starting over? Or should I just forget it and start with fresh materials? What are bitwise operators? I wonder if this is a bug, or if I missed something... When a light hits tangentially convex surfaces with "smooth shading", the shadow on this surface seems to "partially ignore" the smoothness of the surface. Here is an example with a simple cylinder :    On the left, a cylinder with "flat shading". Everything's ok. On the right, the same with "smooth". We can see an edge, bad shadowed.  With or without the "Edge split" modifier, no matters, the edge is still there. The shader is a simple white diffuse shader with no texture nor displace.  I have to add a subsurf modifier with at least 2 levels, which will insanely increase the vertex count... And I can't do that for my project.  I hope someone already saw it, and have a little tip to share. Or if someone already know it's a bug, or a limit of the current Cycles renderer ? What does the "Current event" post notice mean? The recently introduced &quot;Current event&quot;  feels incomplete:  At the very least, it needs a period at the end to complete the sentence. Aside from that, what is a future visitor of a question supposed to take away from this notice? The &quot;citation needed&quot; text explains what expectations are placed on the question bearing the notice and invites action. Is the &quot;current event&quot; notice purely informational? What's the desired effect of adding it to a question? The current on-question notice is:  This post relates to a rapidly changing event.  In addition to answering those questions, I'm hoping we can work out a proposal for a wording change to make it more clear. I can't come up with anything to start with, since I don't quite get what the notice is supposed to achieve in the first place.