5:10 AM
A new tag was created by crystal_math.
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Suppose that $f:[a,b]\rightarrow R$ is continuous and $g:[a,b]\rightarrow R$ is integrable and such that $g(x) \ge 0$ for all $x \in [a, b]$. Prove that there is a number $c$ in $[a,b]$ such that $$\int_{a}^{b}f(x)g(x)dx = f(c)\int_{a}^{b}g(x)dx$$ My proof: Consider $F(x) =f(x)\int_{a}^{b}g(x)... This tag was created and removed before: chat.stackexchange.com/transcript/3740/2018/6/7 Two new tags and were created by ZAF. 0 https://arxiv.org/pdf/1508.07879.pdf Can anybody help me? I don´t understand the proof of the Theorem 2.1 (page 6) The proof is in the page 9. I don´t understand why there exist$L^{-1}$, and I don't understand why$g = PL^{-1}Q$I tried to understand by studying the ore set's theory. But firstly... 3 hours later… 8:01 AM @Brahadeesh My current stance is that I untag "analysis" tag and retag with more suitable tags. If there is objection, I will keep the tag only when the question is about real analysis (or calculus) (similar to what Michael Rozenberg suggested). By doing so I think at least the questions with the "analysis" tags are mainly real analysis questions. user185131 Right, that sounds reasonable. Please continue along those lines :) I’ll also help as much as I can. 8:21 AM @MartinSleziak So the tag was removed and then recreated again: math.stackexchange.com/posts/3661036/revisions Now there are four questions with the tag: 3 For$a,b,c > 0$prove: $$(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}) +\frac{486(ab+bc+ca)^3}{(a+b+c)^6} \geqq 27$$ My work: I can easy found SOS for it: $$\text{LHS-RHS}=\sum {\frac { \left( a-b \right) ^{2}\cdot M}{{a}^{2}{b}^{2} \left( a+b+c \right) ^{6}}} \geqq 0$$ Where$M=\left(...

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For $a,b,c>0.$ Prove$:$ $$\displaystyle \frac{a(b+c)}{a^2+bc}+\frac{b(c+a)}{b^2+ca}+\frac{c(a+b)}{c^2+ab} +\frac{2(a^2+b^2+c^2-ab-bc-ca)}{(a+b+c)^2}+\frac{96(a-b)^2(b-c)^2(c-a)^2}{(a+b+c)^6} \leqslant \frac{(a+b+c)^2}{ab+bc+ca}$$ I found it when I tried to find the stronger version of this, you ...

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For $a,b,c \in (0,\infty).$ Prove$:$ $$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ac}+\frac{c(b+a)}{c^2+ba}\geqq 1+\frac{16abc}{(a+b)(b+c)(c+a)}$$ My proof by SOS$:$ $$\left( {a}^{2}+bc \right) \left( ac+{b}^{2} \right) \left( ab+{c}^{ 2} \right) \left( a+b \right) \left( b+c \right) \left( c+a ... 2 For a,b,c>0. Prove:$${\frac {ab}{ \left( a+b \right) ^{2}}}+{\frac {bc}{ \left( b+c \right) ^{2}}}+{\frac {ac}{ \left( c+a \right) ^{2}}}+\,{\frac { \left( a+b \right) \left( b+c \right) \left( c+a \right) }{16abc}}\geqslant \frac{5}{4}$$AM-GM kills it easy, but I think it's hard to get... @Brahadeesh There are 30K+ questions with the analysis tag..... so help is very welcome (but it also seems that we are in a lost cause) user185131 It's an uphill battle for sure :( I'm not sure what the rate of new [analysis] tagged questions is, but if we can steadily chip away at it then we should be done by... 2050, I guess. :P 9:45 AM Find all n such that 3^{2n+1}+2^{n+2} is divisible by 7 math.stackexchange.com/posts/1604280/revisions The OP writes: "So I am not allowed to use mods, as is a calculus question." Is that sufficient reason to include the [tag:calculus tag]? user185131 9:58 AM @MartinSleziak I’m not sure, the OP has already accepted an answer that just uses induction, so they are probably not looking for a calculus based proof at all. I would prefer not tagging it with [calculus]. Thanks for the response. I have removed the calculus tag - so I am glad that somebody else confirmed this as a reasonable decision. A new tag was created by Rodrigo de Azevedo. Given a collection of points in two, three, or higher dimensional space, a "best fitting" line can be defined as one that minimizes the average squared perpendicular distance from a point to the line. The next best-fitting line can be similarly chosen from directions perpendicular to the first. Repeating this process yields an orthogonal basis in which different individual dimensions of the data are uncorrelated. These basis vectors are called Principal Components, and several related procedures Principal Component Analysis (PCA). PCA is mostly used as a tool in exploratory data analysis and... 0 Can somebody confirm that the number of "samples for each trial" doesn't matter(i guess that's right the language) for PCA. The case at hand is this: i have 5 sets of 3-dimensional datapoints of varying length. So, let's say, (3,330), (3,540), (3,680), (3,1200), (3,2214) in np.shape-speak. Each s... 380 Singular value decomposition (SVD) and principal component analysis (PCA) are two eigenvalue methods used to reduce a high-dimensional data set into fewer dimensions while retaining important information. Online articles say that these methods are 'related' but never specify the exact relation. ... 3 hours later… 1:08 PM FYI: pqr is not a widely accepted term, so it doesn't belong as a tag on this site. — Don Thousand 1 hour ago @DonThousand pqr is an method for symmetric polynomials, same as uvw. — tthnew 1 hour ago @Don Thousand You deleted a tag pqr, which is also uvw. I can show a solution by uvw if you want. — Michael Rozenberg 26 mins ago 1:41 PM @MartinSleziak Apparently these methods apply to symmetric inequalities. Maybe we should create a tag to replace both and (and maybe others)? Personnally I would find such a tag more descriptive. On the other hand, I guess the questions to be tagged like this would be essentially those tagged and , so maybe it is not so useful. -8 I think these tags (uvw and sos) are useful for the forum. User, which looks for to learn these methods, can click these tags and see many examples, how he can prove inequalities by these methods. For example. Let we need to prove that$$a\sqrt{a^2+bc}+b\sqrt{b^2+ac}+c\sqrt{c^2+ab}\geq\sqrt{2(...

The above suggestion has score -8 (no upvotes, 8 downvotes). Despite that, both and were created.
So communication about contest-math-inequalities tags seem to be rather difficult.
In any case, if there are some people who are familiar with the type of questions which are often asked in tag and who understand the tag-system a bit, their input might be useful.

2:09 PM
In my opinion, one of the problems with these inequalities tags is that very often, the same question can be solved using several different known inequalities/methods; since the number of tags on a question is limited, it is often not possible to tag a question with all the relevant solution methods.
By the way, at the moment has been removed from all questions.