[quote="nguoivn"]Given $ a, b, c \geq\ 0$.
Prove that: $ \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca)} \geq\ \frac {a^2}{a^2 \plus{} bc} \plus{} \frac {b^2}{b^2 \plus{} ca} \plus{} \frac {c^2}{c^2 \plus{} ab}$[/quote]
It's equivalent to
$ \sum_{sym}\left(a^5b^3 \minus{} a^4b^4 \plus{} \frac {1}{2}a^6bc \minus{} a^4b^3c \plus{} a^4b^2c^2 \minus{} \frac {1}{2}a^3b^3c^2\right)\geq0,$ which is obvious.
SOS ( $ a^2S_b \plus{} b^2S_a\geq0$ ) helps too but it's very ugly for this nice inequality.

[quote="nguoivn"]Given $ a, b, c \geq\ 0$.
Prove that: $ \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca)} \geq\ \frac {a^2}{a^2 \plus{} bc} \plus{} \frac {b^2}{b^2 \plus{} ca} \plus{} \frac {c^2}{c^2 \plus{} ab}$[/quote]
we used a lemma of VQBC
setting $ p\equal{}a\plus{}b\plus{}c\equal{}1,q\equal{}ab\plus{}bc\plus{}ca\equal{}\frac{1\minus{}t^2}{3}(t\in [0,1]),r\equal{}abc$
$ \Leftrightarrow \minus{}\frac{1}{81}\plus{}\frac{7}{81}t^2\minus{}\frac{5}{27}t^4\plus{}\frac{13}{81}t^6\plus{}\frac{1}{3}r\minus{}\frac{4}{3}rt^2\plus{}8r^2t^2\minus{}\frac{4}{81}t^8\plus{}2rt^4\ge 0$
by lemma we have $ \frac{(1\plus{}t)^2(1\minus{}2t)}{27}\le r\le \frac{(1\minus{}t)^2(1\plus{}2t)}{27}$
*Case 1: if $ f'(r)>0$
$ \Rightarrow f(r)\ge f(\frac{(1\plus{}t)^2(1\minus{}2t)}{27})\equal{}\frac{1}{729}t^2(t\plus{}4)(2\minus{}t)(t\plus{}1)^2(2t\minus{}1)^2\ge 0$
*Case 2: if $ f'(r)<0$
$ \Rightarrow f(r)\ge f(\frac{(1\minus{}t)^2(1\plus{}2t)}{27})\equal{}\frac{1}{729}t^2(t\plus{}2)(4\minus{}t)(2t\plus{}1)^2(t\minus{}1)^2\ge 0$
we have done :P

If $ a,b,c$ be side lengths of a triangle , it has an simple proof :

<=>$ abc(\sum \frac { b \plus{} c \minus{} a}{a^2 \plus{} bc} )\ge \frac {2\sum ab \minus{} \sum c^2}{2}$ ( Note: $ LHS \ge 0 \,forall a,b,c \ge 0$)
Easy to prove :$ \frac { b \plus{} c \minus{} a}{a^2 \plus{} bc} \ge \frac { a \plus{} c \minus{} b}{b^2 \plus{} ac} \,\, if \,b \le a$
By chebysev, we have :
$ abc(\sum \frac { b \plus{} c \minus{} a}{a^2 \plus{} bc} ) \ge \frac {3abc}{a \plus{} b \plus{} c}[(ab \plus{} bc \plus{} ca)( \sum \frac {1}{a^2 \plus{} bc} ) \minus{} 3 ] \ge \frac {9abc(2ab \minus{} \sum a^2)}{(a \plus{} b \plus{} c)( \sum a^2 \plus{} \sum ab )}\ge RHS$
The last ineq is equivalent to Schur with degree 2

About prove that ineq with $ a,b,c \ge 0$
:P . If someone'd like to do it by pqr , this ineq is equivalent to :
$ 2rq(1 \minus{} 3q \plus{} 4r) \ge (4q \minus{} 1)(8r^2 \plus{} r \plus{} q^3 \minus{} 6rq)$
.... It has many ways to prove this ineq :P

I can give a similar example:
If $ a \geq\ b\geq\ c$, we see: $ f(b)\equal{}(b\minus{}a)(b\minus{}c) \leq\ 0$
But if we consider: $ f'(b)\equal{}2b\minus{}c\minus{}a$
If $ f'(b) \geq\ 0$ then $ f(b) \geq\ f(c) \equal{} 0$
If $ f'(b) \leq\ 0$ then $ f(b) \geq\ f(a)\equal{}0$
Therefore, we can say: $ (b\minus{}a)(b\minus{}c) \geq\ 0$     $ ????$
--------------------
I think your proof was wrong same above  :)

[quote="nguoivn"][quote="bunhiacovski"]
*Case 1: if $ f'(r) > 0$
$ \Rightarrow f(r)\ge f(\frac {(1 \plus{} t)^2(1 \minus{} 2t)}{27}) \equal{} \frac {1}{729}t^2(t \plus{} 4)(2 \minus{} t)(t \plus{} 1)^2(2t \minus{} 1)^2\ge 0$
*Case 2: if $ f'(r) < 0$
$ \Rightarrow f(r)\ge f(\frac {(1 \minus{} t)^2(1 \plus{} 2t)}{27}) \equal{} \frac {1}{729}t^2(t \plus{} 2)(4 \minus{} t)(2t \plus{} 1)^2(t \minus{} 1)^2\ge 0$
we have done :P[/quote]
I rather suspect your solution. I think it's not true. Because:
Chẳng hạn với trường hợp đầu tiên, có thể cận dưới này không thuộc khoảng đồng biến của hàm số f(r).
Sory I don't know to say by English. Can some VietNamese translate for me to English ? Thank you :D[/quote]
I think my proof is true because $ t\in[0,1]$, I have another proof but I can not post because it's ugly
[hide]@(-.^)@: em chắc chắn lời giải này ko sai anh àh:D[/hide]

[quote="nguoivn"]I can give a similar example:
If $ a \geq\ b\geq\ c$, we see: $ f(b) \equal{} (b \minus{} a)(b \minus{} c) \leq\ 0$
But if we consider: $ f'(b) \equal{} 2b \minus{} c \minus{} a$
If $ f'(b) \geq\ 0$ then $ f(b) \geq\ f(c) \equal{} 0$
If $ f'(b) \leq\ 0$ then $ f(b) \geq\ f(a) \equal{} 0$
Therefore, we can say: $ (b \minus{} a)(b \minus{} c) \geq\ 0$     $ ????$
--------------------
I think Bunhiacopxki's proof was wrong same above  :)[/quote]

How will you say, arquady ?  :wink:

My proof:
Suppose $ a \plus{} b \plus{} c \equal{} 3$. We need to prove:
$ f(r) \equal{} 4q^4 \minus{} 9q^3 \plus{} 24qr^2 \minus{} 54q^2r \minus{} 72r^2 \minus{} 243r \plus{} 216qr \leq\ 0$
$ f'(r) \equal{} 48qr \minus{} 54q^2 \minus{} 144r \minus{} 243 \plus{} 216q$
$ f''(r) \equal{} 48(q \minus{} 3) \leq\ 0, so f'(r) \leq\ f'(0) \equal{} \minus{} 54q^2 \minus{} 144 \plus{} 216q \leq\ 0$
So, with $ q \leq\ \frac {9}{4}, f(r) \leq\ f(0) \equal{} q^3(4q \minus{} 9) \leq\ 0$
With $ q \geq\ \frac{9}{4}$, we have: $ f(r) \leq\ f(\frac {4q \minus{} 9}{3}) \leq\ 0$ (trues with $ q \geq\ \frac{9}{4}$)

[quote="nguoivn"]Given $ a, b, c \geq\ 0$.
Prove that: $ \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca)} \geq\ \frac {a^2}{a^2 \plus{} bc} \plus{} \frac {b^2}{b^2 \plus{} ca} \plus{} \frac {c^2}{c^2 \plus{} ab}$[/quote]
We have
\[ \sum \frac{2a^2}{(a\plus{}b)(a\plus{}c)} \minus{}\sum \frac{a^2}{a^2\plus{}bc} \equal{}\sum \frac{a^2(a\minus{}b)(a\minus{}c)}{(a\plus{}b)(a\plus{}c)(a^2\plus{}bc)} \ge 0\]
(easy to check by Vornicu Schur)
It suffices to prove that
\[ \frac{(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca)} \ge \sum \frac{2a^2}{(a\plus{}b)(a\plus{}c)} \equal{}\frac{2\sum ab(a\plus{}b)}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\]
Assume that $ a\plus{}b\plus{}c\equal{}1$ and put $ q\equal{}ab\plus{}bc\plus{}ca,r\equal{}abc$, then the inequality becomes
\[ \frac{1}{4q} \ge \frac{q\minus{}3r}{q\minus{}r}\]
\[ \Leftrightarrow \frac{q\minus{}r}{q\minus{}3r} \ge 4q\]
\[ \Leftrightarrow \frac{2r}{q\minus{}3r} \ge 4q\minus{}1\]
By Schur's Inequality for third degree, we have $ r \ge \frac{4q\minus{}1}{9}$, then
\[ \frac{2r}{q\minus{}3r} \ge \frac{2r}{q\minus{} \frac{4q\minus{}1}{3}} \equal{}\frac{6r}{1\minus{}q}\]
It suffices to show that
\[ 6r \ge (4q\minus{}1)(1\minus{}q)\]
But this is just Schur's Inequality for fourth degree
\[ \sum a^4\plus{}abc\sum a \ge \sum ab(a^2\plus{}b^2)\]
We have done. :)

Yes, I very like can_hang's proof  :lol:

[quote="mitdac123"]If $ a,b,c$ be side lengths of a triangle , it has an simple proof :

<=>$ abc(\sum \frac { b \plus{} c \minus{} a}{a^2 \plus{} bc} )\ge \frac {2\sum ab \minus{} \sum c^2}{2}$ ( Note: $ LHS \ge 0 \,forall a,b,c \ge 0$)
Easy to prove :$ \frac { b \plus{} c \minus{} a}{a^2 \plus{} bc} \ge \frac { a \plus{} c \minus{} b}{b^2 \plus{} ac} \,\, if \,b \le a$
By chebysev, we have :
$ abc(\sum \frac { b \plus{} c \minus{} a}{a^2 \plus{} bc} ) \ge \frac {3abc}{a \plus{} b \plus{} c}[(ab \plus{} bc \plus{} ca)( \sum \frac {1}{a^2 \plus{} bc} ) \minus{} 3 ] \ge \frac {9abc(2ab \minus{} \sum a^2)}{(a \plus{} b \plus{} c)( \sum a^2 \plus{} \sum ab )}\ge RHS$
The last ineq is equivalent to Schur with degree 2

[/quote]

I checked your proof and I found it is wrong. If you apply shur at your last expression in the form

$ \frac{9abc}{\sum a} \geq 2\sum ab \minus{} \sum a^2$

you obtain

$ \sum a^2 \leq \sum ab$ surely false. :wink:

[quote="nguoivn"]Given $ a, b, c \geq\ 0$.
Prove that: $ \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca)} \geq\ \frac {a^2}{a^2 \plus{} bc} \plus{} \frac {b^2}{b^2 \plus{} ca} \plus{} \frac {c^2}{c^2 \plus{} ab}$[/quote]
The following inequalities are true too.
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
1.
\[ \frac {a \plus{} b \plus{} c}{2\sqrt [3]{abc}} \geq\ \frac {a^2}{a^2 \plus{} bc} \plus{} \frac {b^2}{b^2 \plus{} ca} \plus{} \frac {c^2}{c^2 \plus{} ab}
\]
2.
\[ \frac {5}{2} \minus{} \frac {ab \plus{} ac \plus{} bc}{a^2 \plus{} b^2 \plus{} c^2} \geq\ \frac {a^2}{a^2 \plus{} bc} \plus{} \frac {b^2}{b^2 \plus{} ca} \plus{} \frac {c^2}{c^2 \plus{} ab}
\]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=246653

to 2]

$\leftrightarrow$

$\sum \frac{2ab}{c^2+ab}\ge \frac{(a+b+c)^2}{a^2+b^2+c^2}$

$LHS \ge \frac{2(\sum ab)^2}{(\sum ab)^2-abc(a+b+c)}\ge \frac{(a+b+c)^2}{a^2+b^2+c^2}$

$\leftrightarrow$
$abc(\frac{a^2+b^2+c^2}{ab+bc+ca}+2)^2\ge (a+b+c)(2(ab+bc+ca)-a^2-b^2-c^2)$

but this follows immediately from schur's inequality in 3-degree
$9abc\ge (a+b+c)(2(ab+bc+ca)-a^2-b^2-c^2)$

[quote="arqady"]
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
1.
\[ \frac {a \plus{} b \plus{} c}{2\sqrt [3]{abc}} \geq\ \frac {a^2}{a^2 \plus{} bc} \plus{} \frac {b^2}{b^2 \plus{} ca} \plus{} \frac {c^2}{c^2 \plus{} ab}
\]
[/quote]
Here is my proof from 2008.
Let $a=\min\{a,b,c\}$, $f(a,b,c)=\frac {a \plus{} b \plus{} c}{2\sqrt [3]{abc}}-\sum_{cyc}\frac {a^2}{a^2 \plus{} bc}$ and $abc=1$.
Hence, $f(a,b,c)-f\left(a,\sqrt{bc},\sqrt{bc}\right)=$
$=\left(\sqrt b-\sqrt c\right)^2\left(\frac{1}{2}+\frac{a\left(\sqrt{bc}-a\right)(b+\sqrt{bc}+c)^2}{(a+\sqrt{bc})(b^2+ac)(c^2+ab)}\right)\geq0$.
Thus, it remains to check our inequality for $c=b$ and $a=\frac{1}{b^2}$, which gives
$(b-1)^2(2b^{10}-2b^8-b^7+b^5+5b^4+3b^3+b^2+2b+1)\geq0$, which is obvious.
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=335838

[quote=nguoivn]Given $ a, b, c \geq\ 0$.
Prove that: $ \frac{(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca)} \geq\ \frac{a^2}{a^2\plus{}bc} \plus{} \frac{b^2}{b^2\plus{}ca} \plus{} \frac{c^2}{c^2\plus{}ab}$[/quote]

[img]http://s16.sinaimg.cn/middle/006ptkjAzy76wc1XrUHdf&690[/img]

[quote=nguoivn]Given $ a, b, c \geq\ 0$.
Prove that: $$ \frac{(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca)} \geq\ \frac{a^2}{a^2\plus{}bc} \plus{} \frac{b^2}{b^2\plus{}ca} \plus{} \frac{c^2}{c^2\plus{}ab}$$[/quote]
Let $a,b,c$ are positive reals . Prove that$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ca}+\frac{c^2}{c^2+ab}<2.$$

[hide=Thanks.][quote=ytChen][quote=sqing]Let $a,b,c$ are positive reals . Prove that$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ca}+\frac{c^2}{c^2+ab}<2.$$[/quote]

[b]Solution[/b]. By the Cauchy-Schwarz's Inequality we have that, for any positive real numbers $a,b,c$,
\begin{align*}&\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ca}+\frac{c^2}{c^2+ab}\\
=&3-\left[\frac{bc}{a^2+bc}+\frac{ca}{b^2+ca}+\frac{ab}{c^2+ab}\right]\\
=&3-\left[\frac{(bc)^2}{a^2bc+(bc)^2}+\frac{(ca)^2}{b^2ca+(ca)^2}+\frac{(ab)^2}{c^2ab+(ab)^2}\right]\\
\le&3-\frac{(bc+ca+ab)^2}{a^2bc+b^2ca+c^2ab+(bc)^2+(ca)^2+(ab)^2}\\
=&2-\frac{a^2bc+b^2ca+c^2ab}{a^2bc+b^2ca+c^2ab+(bc)^2+(ca)^2+(ab)^2}\\
<&2,
\end{align*}where the last inequality holds since $a^2bc+b^2ca+c^2ab>0$ for any $a,b,c>0$. $\blacksquare$[/quote]

[/hide]
Another proof：


topology
real analysis
real analysis theorems

These are some of my practice problems:
1. Show a set $F_{A}\subseteq A \subseteq X$ is closed subset of A iff there exists a closed subset $F_{X}$ of $X$ such that $F_{A}= F_{X}\cap A$.

2. Supposed $A \subseteq B \subseteq X$. Show that if $A$ is open in $B$ which is open $X$, then $A$ is open in $X$.

I wasn't sure if one could do 2 as follows:
Choose $a \in A$. There exists $\frac{r}{2}> 0$ such that $d(a, b) < \frac{r}{2}\Rightarrow b \in A$ and for each $b \in B$ we have $d(b, x) < \frac{r}{2}\Rightarrow x \in B$.
We have $d(a, x) \leq d(a, b)+d(b, x) < \frac{r}{2}+\frac{r}{2}= r \Rightarrow x \in A$. So that $A$ is open in $X$.

These are utterly trivial if we use the general definitions of the relative topology: $U$ is open in $A\subset X$ when there is some open $G\subset X$ with $U=A\cap G$.

1. Let $G_{A}=A\setminus F_{A}$; $G_{A}$ is open in $A$ iff there is some open $G_{X}\subset X$ with $G_{A}=A\cap G_{X}$. If such a set exists, let $F_{X}=X\setminus G_{X}$. If a $F_{X}$ exists, let $G_{X}=X\setminus G_{X}$.

2. $A=G\cap B$ for some open $B$. The intersection of two open sets is open.


topology
geometry
geometric transformation
invariant
advanced fields
advanced fields unsolved

Let $ G$ be a compact Hausdorff topological group. Prove that $ G$ is finite or uncountable.

use the normed haar measure $ \mu$ on $ G$. if $ G$ was countable, say $ G \equal{} \{g_1,g_2,...\}$, we have $ 1 \equal{} \sum_{i \in \mathbb{N}} \mu(\{g_i\})$. by the fact that $ \mu$ is translation-invariant, all these singletons have the same measure, a contradiction.

there's a more elementary way: every perfect set (every point is an accumulation point) is uncountable.

and, in fact, every infinite compact topological group is perfect.


For positive integers $n$, define $f(n)=1^n+2^{n-1}+3^{n-2}+\cdots+(n-2)^3+(n-1)^2+n$.

What is the minimum value of $\frac{f(n+1)}{f(n)}$?

This isn't so bad:
[hide]Computing the first few values by hand, we get 1, 3, 8, 22, 65, 209, 732, with ratios that start at 3, drop to 8/3, and then climb, eventually passing 3.  Let's consider, for "large enough" $n$, the expression $f(n + 1) - 3f(n)$.  If we subtract corresponding terms, this gives us $-2 - 2^{n - 1} + 4^{n - 3} + 2\cdot 5^{n - 4} + \ldots + n(n - 3) + (n +1)$.  When $n > 3$, all terms but the first two are positive, and when $n > 5$, the third term $4^{n - 3} = 2^{2n - 6} > 2^{n - 1} + 2$ so the whole expression is positive and $\frac{f(n + 1)}{f(n)} > 3$.  Combined with our data for $n \leq 6$, this shows the absolute minimum is 8/3, achieved when n = 2.[/hide]

The (I think) natural question this raises: show that the given ratio approaches infinity as n grows large, and (if possible) say about how quickly.


Well I'm new here... as you can probably tell.

In most other forums I've visited, there has been an Introductions section so people can get to know each other. It would probably be a good idea to have one here as well. Then again, I'm just a noob, so who cares what I think?

Maybe...
But there are over 40000 users here so I don't know.

Well, there are such topics in the FF, but they eventually "die-out"

http://www.mathlinks.ro/viewtopic.php?t=15747

http://www.mathlinks.ro/Forum/viewtopic.php?t=20070


A chesstable coloured black-white has the dimensions $ m\times n$,where $ m,n$ are naturals.On this table are made $ moves$ of this type: There are taken two fields $ 1\times1$ which have a common side and modify their colours as it follows:
[i]white[/i] field becomes [i]black[/i],the [i]black [/i]field becomes [i]red[/i] and the [i]red [/i]field becomes [i]white[/i].
Find all the possible values for $ m$ and $ n$ s.t. all the initial white fields become black and
all the initial black fields become white.

How is it coloured initially? I will assume that the initial colouring is like the common chessboard.

So, we take three consecutive squares, WBW (white black white). First, take the first two, so we have BRW. Then take the last two, so we have BWB. We can do this for any three consecutive squares, and change all black squares into white and vice versa.

Therefore, I think the initial condition should be $ 3|mn$, i.e. $ m$ or $ n$ is divisible by $ 3$.

Answer to your question:The initial colouring is like the common chessboard.

You proved that $ m \equal{} 3k$ and any $ n$ natural and viceversa satisfies the condition.

But how you exclude other cases, for example $ m \equal{} 3a \plus{} 1, n \equal{} 3b \plus{} 1$? :?:

actually it can be proved very easily (as i did today by congruences)
n(w)=number of initial white
n(b)=number of initial black
first we observe that number of transformations on initial white t(n(w))= number of tranformations on initial black t(n(b))
now also n(w)=t(n(w))mod 3
and 2n(b)=t(n(b))mod 3
subtracting we get n(w) - 2n(b)=0 mod 3 ( because t(n(w))=t(n(b)))
 now we add 3n(b) to RHS
we get n(w) + n(b)=0mod3
n(w) + n(b)=total number of fields = mn
therefore 3 divides mn
hence done


My document starts with \documentclass[11pt]{article} . I've tried changing that 11pt to another value: I want it to have a minimum size of 18pt; but 12pt is greater than 26pt type or anything larger than 12pt. (Also, when I compile, it says "Unused global option: [26pt]"). \large, \Large, \LARGE, etc. don't work because they don't change the heading sizes.

Also how do I decrease the margins?

Hi,
If you want to change the font size of the chapter/section headings (I couldn't make out exactly what it was that you needed), an easy way to do it is to use the titlesec package. For example, for Large section titles, use:
[code]
\titleformat{\section}[hang]{\center\Large}{\Large\thesection}{12 pt}{}{}
[/code]
A simple way to increase margins is the fullpage package (I think it shrinks the margins to 1'' all around). If you want more control, try this: [url=http://www.image.ufl.edu/help/latex/margins.shtml]margins[/url].

SSN.

I have discovered that if you use \documentclass{memoir}, you can change the font size to 10,11,12,14, or 17pt. In case anyone else runs into this problem. \usepackage{scalefnt}...\scalefont{1.4} also works, but doesn't affect the headings, which is annoying.

The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=extsizes][b]extsizes package[/b][/url] give you a range of 8-20pt for the base size, works well and is easy to use.


if \[ac=b^{2}\] , \[a+b+c=3\] , find the min and max of the \[b\]

If $b>0$, then:
$3=a+b+c \geq 2\sqrt{ac}+b = 2b+b=3b$. Therefore $b \leq 1$ so $b_{\text{max}}=1$.

If $b<0$, then:
$3=a+b+c \geq 2\sqrt{ac}+b =-2b+b=-b$. Therefore $b \geq-3$ so $b_{\text{min}}=-3$.

that is say the answer of my book is wrong
thank u  :wink:


1. 2-bromoethanol and 1,2-dibromomethane
2. 4-chloro-1-butene and n-butyl chloride
3. 1-hexene and 2-hexanol
4. 1-chloro-2-methyl-2-propanol and 1,2-dichloro-2-methylpropane

  [hide="i think"]
1.chromic anhydride in sulfuric acid
2.bromine carbon-tetrachloride
3.bromine carbon tetrachloride/chromic anhydride in sulfuric acid
4.chromic anhydride in sulfuric acid[/hide]

Solubility tests could also be used. In 1., 2-bromoethanol is soluble in water, while 1,2-dibromoethane isn't. In 2., 4-chloro-1-butene is soluble in sulphuric acid, while n-butyl chloride is not. For 3., the alcohol should be more soluble that the alkene. And in 4., the alcohol is more soluble than the alkyl halide.

Infrared spectroscopy could also be easily used to distinguish each pair of compounds:

1. 2-bromoethanol would show the broad band from the O-H centered at about 3200-3400 cm-1, while 1,2-dibromoethane wouldn't have any band in that region.

2. 4-chloro-1-butene has a vinyl group and so we could find the following bands: alkene C-H stretch at about 3070 cm-1 (medium-strong); overtone of the ~920 cm-1 band at about 1840 cm-1 (weak); two strong bands for alkene C-H bending vibrations, one at about 990 cm-1, and another at 920 cm-1. n-Butyl chloride would have naturally these bands.

3. 1-Hexene would show the vinyl bands I referred above. 2-hexanol, would show the broad O-H band at ~3300 cm-1, and also the C-O stretching band (strong) at about 1100 cm-1 (because it is a secondary alcohol).

4. 1-chloro-2-methyl-2-propanol would also show the broad O-H band at ~3300 cm-1, but now we would find the C-O band at about 1150 cm-1, because we have here a tertiary alcohol.


I want to create my own theorem-like environment, since I don't like the numbering and fonts used in the default one.

The default LaTeX "newtheorem" creates an environment that puts some space before and after each theorem to make the theorem stand out.

So I did the following to mimic the environment
[code] \newenvironment{prop}[1]{\bigskip \noindent {\bf Proposition #1.}}{\bigskip} [/code]
This is perfect when I have
[code]
Paragraph

\begin{prop}{1}
Statement
\end{prop}

Paragraph
[/code]
However, when I do the following
[code]
Introducing two propositions:

\begin{prop}{1}
Statement1
\end{prop}

\begin{prop}{2}
Statement2
\end{prop}
[/code]
I get twice the space between the two propositions.

How do I make the environments so that it onlys puts a certain amount of space before and after, such that writing one such environment after the other does not double the space?

Or is there a better way to do this?

Well, somehow figuring out how to change the settings on the original would be ideal, but neither of us knows how to do that right now. The easiest thing i can think of would be to make a second environment that is identical to the first but doesn't have the initial bigskip:
[code]\newenvironment{propfollow}[1]{\noindent {\bf Proposition #1.}}{\bigskip}[/code]
Hope this helps/works!

Or, you could give the environment more optional parameters, so you can explicitly provide the before and after spacing each time you use it. However, since you'll be using it with a bigskip before and after almost every time, this might get cumbersome. The second environment for exceptional cases is probably easier.

Oh, btw, you could use counters to automatically number the theorems instead of doing that manually. You would do (something like) this:

\newcounter{probnum} (or whatever else you want to call it)
\setcounter{probnum}{1}

\newenvironment{prop}{
\bigskip \noindent {\bf Proposition \theprobnum.}
\stepcounter{probnum}
}{\bigskip}

These don't like to go in the preamble (I don't know why). Put them first thing after \begin{document}.

Of course, this simply counts 1, 2, 3, ..., but more complicated stuff can be done if necessary. (And, if you use the propfollow idea from before, the same counter can be used, and it will number in order as necessary.)

I think I found a possible solution.

The command \addvspace{12pt} adds the space, but does not add more space if there is already that much space in place.


logarithms
function
inequalities
number theory unsolved
number theory

Do you mean for all $ n > 1$? Or did you mean $ \sigma(n) \leq n\sqrt{n}$? What you're saying isn't true when $ n\equal{}1$, as $ \sigma(1) \equal{} 1 \equal{} 1\sqrt{1}$.

False when $ n \equal{} 2$.

When $ n > 2$, we can show that $ n$ has at most $ 2 \sqrt {n} \minus{} 1$ divisors and of those divisors all except $ 1$ are greater than or equal to $ 2$.  Now, $ \frac {\sigma(n)}{n} \equal{} \sum_{d | n} \frac {1}{d} \equal{} 1 \plus{} \sum_{d | n, d > 1} \frac {1}{d} < 1 \plus{} \frac {\sigma(n) \minus{} 1}{2} \le 1 \plus{} \frac {2 \sqrt {n} \minus{} 2}{2} \equal{} \sqrt {n}$.  This bound is tightest when $ n \equal{} 3, 4$.

The above argument suggests the following tighter bound:  $ \frac {\sigma(n)}{n} < H_{ 2 \sqrt {n}} \sim \frac {1}{2} \ln n$.  In fact, we can do [url=http://en.wikipedia.org/wiki/Divisor_function#Approximate_growth_rate]even better[/url].

[b]Edit:[/b]  Follow-up.  Show that $ \frac {\sigma(n)}{n}$ is unbounded.

[b]Edit 2:[/b]  [hide="Discussion"] Letting $ n \equal{} \prod p_k^{e_k}$, we can find that

$ \frac {\sigma(n)}{n} \equal{} \prod \left( 1 \plus{} \frac {1}{p_k} \plus{} ... \plus{} \frac {1}{p_k^{e_k}} \right)$.

Let $ n_m \equal{} p_1 p_2 ... p_m$ be the primordials.  Then

$ \frac {\sigma(n_m)}{n_m} \equal{} \prod_{k \equal{} 1}^{m} \left( 1 \plus{} \frac {1}{p_k} \right)$,

the natural log of which is bounded by $ H_{m}$ (or something along those lines).  Some more arguments along these lines should give the growth rate in the Wikipedia article.  (Note that as $ m$ increases without bound the product on the RHS diverges.) [/hide]

I'm sorry. Right with n>2. Thanks your  solution

See also [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=83692]An inequality on the sum of divisors.[/url]


Prove that:
$\sqrt[3]{\frac{x}{x+26}}+\sqrt[3]{\frac{y}{y+26}}+\sqrt[3]{\frac{z}{z+26}}\geq 1$
where $x;y;z$ are positive real numbers satisfying $xyz=1$


Here's a thread for discussion of the OCTM test.  So, which problems did you have trouble with, and how do you think (or know) you did?

I thought it seemed easy overall, the only ones I didn't answer were 22 and 26.  I did catch 2 or 3 mistakes on the way home, so I think I got like 35, but I'm very happy with that, assuming it's correct (I only got 24 last year...)

Btw, could someone post the answers for it?  I didn't get an answer sheet after I took it, because I was taking it at an irregular location due to an In the Know tournament...

I was able to answer all the questions. Unfortunately I got the box and whiskers graph one wrong. I was really upset to ever see that thing again (I always complain about how I didn't learn anything in 6th grade except that useless graph, but I can't call it useless any more :mad: I should've paid more attention...)

I had just corrected an answer when the proctor called time... man, that was soo scary, i barely made it. Whew.

Lol, I feel for you man. I HATE box and whisker plots with a passion.

DANGIT i was thinking of SSA, its not AAS...

and isn't the average speed the total distance divided by total time?

I interpreted it as the average of the two rates separately. I found out $r_{1}$ and $r_{2}$.

hmm.. the question was kind of ambiguous. By the way "average speed" is defined, my way should be correct I believe, but your answer is more like the kind of thing the question would ask for.

[quote="tjhance"][quote="roadnottaken"]The OCTM used Sudoku for one of its problems! :rotfl:  :rotfl:

I took last year's OCTM as practice for this year's test; this one was decidedly easier.  I worked out all of the problems except #22, #26, and probably #36 (can't remember for certain) during the test.  Afterwards, I bludgeoned #26 to death with the laws of sines and cosines until it yeilded the value of y, but when I found out that the formula for 22 was just equivalent ratios, I wanted to scream!

I also caught a few mistakes while typing up the following list:

Optimistic prediction: 35
Unduly pessimistic prediction: 30

[hide="I think the correct answers are:"]
1.  $2005$
2.  $a)$ $f;$ $b)$ $f;$ $c)$ $t$
3.  $side$
4.  $2$ $yd,$ $1$ $ft,$ $9$ $in$
5.  $AAA$
6.  $55\%$
7.  $876495$
8.  $C$
9.  $-130$
10. $1$
11. $\frac{17}{20}$
12. $b=1$, $c=-1$, $d=-12$
13. $-\frac{4\sqrt7}7$
14. $A$
15. $XVII$
16. $3+2\sqrt3$
17. $10300$
18. $1.5$
19. $40$
20. $all$ $real$ $numbers$
21. $29.5\%$
22. $x=3\frac37$
23. $\frac12$
24. $D$ and $F$
25. $\frac{12}{17}$
26. $4.4641$
27. $5040$
28. $2\frac13$
29. $E$
30. $y=22.5$
31. $10$
32. $6$
33. $\$12.10$
34. $-243$
35. $3^{\circ}59'$
36. $x=\frac{2\sqrt{10}}{5}$
37. $12$ $in$
38. $slide$ $rule$
39. $\sqrt{29}$
40. $126$ $mph$[/hide][/quote]

[hide="a few errors"]
5. AAS is also  invalid theorem.
7. 876593 is higher
12. I got $b=-1,c=-12,d=0$
17. did you mean 10200?
24. shoot you're right, i'm wrong
40. $\frac{280+280}{2+2.5}=124\frac{4}{9}$
[/hide][/quote]
[hide="@tjhance"]7.  Oops... not that it affects my score; my original answer was $876594$, which is even.
12.  That was an error in copying from my test sheet to this topic; I had it right on the contest.
17.  Yes[/hide][hide="@Xaero"]
34. How did you get [i]that[/i]?
36. Heh.  I think I completely missed that problem during the contest, anyway, but you are right.
37. Arrgh!  I thought it was volume.[/hide]Revised estimate: 33/34 :maybe:

[quote="Xaero"]I think I got a 37 if I didn't copy the answers down wrong. Yeah, I killed problem 26 with Stewart's Theorem.  :rotfl:  :D[/quote]

man, i love that theorem.

i thought it was weird that they asked for the name of the slide rule... kind of an odd question. Even though those things are like the most beastly computation tool ever created.

It took me awhile to do the Roman Numeral one... I had to remember exactly how it worked. I had to come back later.

I was worried about 20, cause i wasn't sure whether "union" was an OR operator or an AND operator... luckily i correctly reasoned it's an OR  :D

I'm curious, what did you guys get for 32? I got $24$ as the answer.

I think you either confused inscribed with circumscribed or calculated the perimeter rather than the side length.  I got $6$ on that one.  See my answers, above.

It is definitely $24$ for number 32.
Oh, and @roadnottaken:
[hide="34"]You added $(2\sqrt{6})^{2}(27)$ rather than subtracting it...[/hide]
I made a bunch of extremely stupid mistakes, which I think is mainly due to lack of sleep. I had a thing going on Friday night, and I didn't get home until 2:30, then I had to get up at 4:30 to make it to my testing site on time. Oh, well, I'll just have to ace it next year.

Revised estimate: 30 (Really hope its enough to make OHMIO)

Do you guys know where I might be able to find a copy of this test, or when a copy might be put up online by? Thanks.

[quote="Xaero"]The test and solutions are now up on the OCTM website.

[url]http://www.octmtournament.org[/url][/quote]

darn, I got the last one wrong... I still think it wasn't worded very well.

about what score is usually required to proceed to OHMIO?

My score: 32

I made way to many mistakes from not reading questions carefully enough or through sheer carelessness.
 :wallbash: $\ldots$ :wallbash_red: $\ldots$ :stretcher: $\ldots$ [huge doctor's bill]

It turns out that the answer for 40 was indeed the average of the two rates.

My score: 37

Thinking about the last question, you took into account the wind speed with the plane speed when it just asks for the plane air speed. There is a difference.

Also,  OHMIO's cutoff is usually around 30.

[quote="Xaero"]Also,  OHMIO's cutoff is usually around 30.[/quote]

I think i should make it then; I made some lame-o mistakes, but not too many.

[quote="mathgeniuse^ln(x)"]Anyone got the contest questions themselves.[/quote]

[url]http://www.octmtournament.org/octmattach/testresults/000000000057.pdf[/url]

Wow, mine was riddled with absolutely retarded mistakes.  From what should have been a 35, I got a 27...

Note to self: get more than 2 hours of sleep before OCTM next year...

[quote="life=tennis+math"]I just took the test (I don't live in Ohio, so I just used the questions from online) and did the test under general conditions (except maybe less adrenaline/pressure). I got a 34, how would that stand in Ohio?
[hide="mistakes"]
8. Dumb mistake...
15. Completely forgot Roman numerals, made a wild guess
20. Forgot what union meant so guessed, picked the wrong one.
30. Eh I don't know what I did
34. Ditto as 30
38. No clue at all. [/hide][/quote]

I'd be happy to answer any questions if anyone needs to.  I got 39.

And 38 is Napiers Bones if anyone cares.  It is an eponym

[quote="mathgeniuse^ln(x)"]I'd be happy to answer any questions if anyone needs to.  I got 39.

And 38 is Napiers Bones if anyone cares.  It is an eponym[/quote]

Nice job on the 39.  That might be 1st in the state...

Hmmm, Napier's Bones.  We did a thing on that in math club at my school last year.  It's pretty cool...

The correct answer was slide rule though.  The site told me...

Number 38 was a "history of math" question, of which they have one every year.  You can not really pin down a difficulty level for those questions; either you know it or you don't.  I happened to know what a slide rule was, but generally I don't know those answers.  It's really just luck.

When I saw that question on the test, I thought it was really funny, remembering one of my friends from another math forum (much smaller and having only been running for about three months) who used a slide rule on the OCTM test when he took it (not because he is that old).

Also, that 39 probably is not best in the state, simply because mathgenius^ln(x) is not in the state (he's from Michigan).

Oh, well, that would prevent that, wouldn't it?  Maybe it's Xaero then.  The top is usually about 37-39...

I highly doubt that I got the #1 score in Ohio since the state champion last year was a 9th grader because he score a 39...
Oh well, I can still hope for top 10 or something.

[quote="mathgeniuse^ln(x)"]Yeah, the 39 includes the fact that I got #38 wrong, but I still think Napier's Bones is an allowed answer.

Yeah, I am from Michigan.[/quote]
Is Napier's Bones another name for a slide rule?

I don't know whether they would have allowed it, but normally they list all acceptable answers in their answers, and the only one for that question was "slide rule".

Well, the slide rule is a outgrowth of Napier's Bones, so I don't know.

Here is a link.

[url]http://www.maxmon.com/1600ad.htm[/url]

In that case, I have a second reason why Napier's Bones is probably not an allowed answer.  The description in the OCTM clearly showed that it was referring to an instrument with moving parts.  Napier's Bones does not fit that description at all.

Oh, I didn't see that.

NVM

It is sliderule, my second answer.


Let a triangle $ ABC$ with Euler line cut $ AB,AC$ at $ M,N$. Prove that: Euler line of triangle $ AMN$ is paralell with $ BC$.

See here for a slightly more general statement (though equivalent afterall): [url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1701732746&t=191643[/url].