let a,b,c>0 such that min{a;b;c}$ \geq\frac{a\plus{}b\plus{}c}{5}$    Prove  that
$ 2(\frac{a\plus{}b}{c}\plus{}\frac{b\plus{}c}{a}\plus{}\frac{c\plus{}a}{b}) \geq \frac{ab}{c^2}\plus{}\frac{bc}{a^2}\plus{}\frac{ca}{b^2}\plus{}9$
This  inequality is true sure and it is very easy  HAHAHA

I don't think this inequality is wrong.
let x=$ \frac{b}{c}$; y=$ \frac{c}{a}$; z=$ \frac{a}{b}$; we have
xyz=1 we need to prove the inequality
2(x+y+z+xy+yz+zx)$ \geq xy^2\plus{}yz^2\plus{}zx^2\plus{}9$
Let t=x+xy-1;u=y+zy-1;v=z+xz-1
denote min{a;b;c}$ \leq\frac{a\plus{}b\plus{}c}{5}$ ,we have t=$ \frac{b}{c}\plus{}\frac{b}{a}\minus{}1 \geq \frac{4b}{a\plus{}c}\minus{}1 \geq0$;u$ \geq0;v\geq0$
With xyz=1 we have tu+uv+vt+uvt=4
and we need to prove :t+u+v$ \geq tu\plus{}uv\plus{}vt$
Let p=t+u+v;
      q=tu+uv+vt
       r=tuv
 We have p+r=4:$ p^2\geq 3q$
By Schur we have: $ p^3\plus{}9r \geq 4pq$
                             $ p^2q^2\plus{}p^3r \geq p^2q^2 \plus{}\frac{9q^2r}{p} \geq 4q^3\equal{}q^4\plus{}q^3r$
                              $ q^2(p^2\minus{}q^2)\plus{}r(p^3\minus{}q^3)\geq 0$
                              $ p \geq q$
Done :rotfl:


In how many ways 23 distinct balls can be partitioned into lots of 4, 4, 5 ,5 and 5 exactly?

[quote="pankajsinha"]$ \frac {(23)!}{(4!)^2 (5!)^3 (2!)(3!)}$[/quote]

Pankaj, Could you please explain how did u get this

It is just to divide the extra-counting which occurs since the changing of ball positions in the same group doesn't count twice
and so does the changing of 4-group or 5-group positions.  :D


For all positive integers $n$, define $a_n = 0$ if $n$ has an even number of distinct prime divisors and $a_n = 1$ otherwise.
Is the number $0.a_1a_2a_3\cdots$ rational or irrational?

[hide]Notice that only perfect squares have an odd amount of factors because the square root of the perfect square is multiplied with itself to create the square.  All other numbers have an even amount of factors.  Our decimal is 0.100100001...  Obviously, the 1s are separated by increasingly large amounts of 0s, so this isn't rational because there's no fixed repeating period.[/hide]

That's wrong. We're talking about *distinct prime divisors* here.

I'm not sure how rigorous of a solution you are looking for, but very informally:
It is well known that the primes are scattered about the integers in a way that defies any simple pattern.
Likewise, the integers that can be expressed as the product of two primes are not periodic.  We therefore have an infinite number of 1s and 0s that can not be determined by some linear function of n (or any simple function of n for that matter).  Therefore,  $a_n$ cannot become cyclical.

Olympiads look for more rigorous solutions.  Here is a sketch of how I think the proof should go:

Assume $0.a_1a_2a_3\cdots$ is rational.
Then, since the digits cannot terminate (since there are infinitely many primes and therefore infinitely many 1s in the decimal expansion above), we conclude that they repeat after some value k.  Let the period be r and let $p$ be the first prime $\geq  k$.
Write a whole period of digits:
 $a_p = 1, a_{p+1}, a_{p+2}, \cdots, a_{p+r}$, and $a_{p+r} = a_p = 1$

Now consider the sequence
$a_{2p} = 0, a_{2p+1}, a_{2p+2}, \cdots, a_{2p+r}$.

Can we get a contradiction?

I think we can show that the primes alone (who have an odd number of prime divisors) are enough to cause a contradiction.

Specifically, by Dirichlet's Theorem on Arithmetic Progressions I think we can show that any attempt to find a period in $a_n$ results in the contradiction that all members of the period must be $1$.

notation: instead of $a_n$, i will use $f(n)$ since it latexes better

theorem: a number is rational iff it can be written as a repeating decimal (not including the first few digits)

assume for sake of contradiction, that the sequence is rational, then the decimal repeats with a period of $k$, suppose that this block of digits begins after $m$ digits

then we have that

$f(m)=f(m+ak)$, $f(m+1)=f(m+ak+1)$, etc where $a$ is any positive integer

$m=0$

look at $f(k)$ and $f(pk)$, where $k$ is a positive integer, and $p$ is a prime that is not in the factorization of $k$,

$k$ is finite (for it to have a period), and primes are infinite, there must be some prime that is not in the factorization of $k$, then this is another prime factor that changes the partity of $f(k)$ and $f(pk)$, so contradiction...

$m\in\mathbb{Z}^+$

we want to show that this is false for sake of contradiction: $f(m)=f(m+ak)$
let $a=m*q$ where $q\in\mathbb{Z}^+$

then $f(m)=f(m\cdot[1+qk])$

by dirchlet's theorem, there are an infinite number of primes in the form $1+qk$, and $m$ must be finite for the sequence to make a rational number, yet there are an infinte number of primes that $1+qk$ can take on, so using one of those primes not in m, changes the partity, thus those terms are not equal, contradition

the expansion is [b]irrational[/b]!

Indeed. Can you also prove it without using Dirichlet?

I also used it by using Dirichlet's theorem ... is there a solution without it ?

i am quite suprised that there would be a proof without using dirchlet's theorem because that would mean that there exists an easy proof for it using the result of this problem  :?

No, it wouldn't prove the Dirichlet theorem in an easy way, and yes, there is a proof which doesn't use Dirichlet.

Oh well, maybe I should just post it.

Suppose that our number is rational and denote $a_n = f(n)$. There exists a positive integer $M$ and a positive integer $a$ (the period) such that for all $x > M$, we have $f(x) = f(x + a)$. Choose $t > 0$ such that $a^t > M$. We have $f\left(a^t\right) = f\left(a^t + na\right)$ for all positive integers $n$. Now choose $n = (p - 1) \cdot a^{t - 1}$ where $p$ is a prime number which does not divide $a$. Then $f\left(a^t\right) = f\left(p \cdot a^t\right)$, and that's a contradiction. $\blacksquare$


A positive integer $d$ is called [i]nice[/i] iff for all positive integers $x,y$ hold: $d$ divides $(x+y)^{5}-x^{5}-y^{5}$ iff $d$ divides $(x+y)^{7}-x^{7}-y^{7}$ .
a) Is 29 nice?
b) Is 2006 nice?
c) Prove that infinitely many nice numbers exist.

First lets note that \[(x+y)^{5}-x^{5}-y^{5}=5xy(x+y)(x^{2}+xy+y^{2})\] and \[(x+y)^{7}-x^{7}-y^{7}=7xy(x+y)(x^{2}+xy+y^{2})^{2}.\] We will show that if $p$ is a prime and $p\not=5,7$ then $p/(x+y)^{5}-x^{5}-y^{5}$ iff $(x+y)^{7}-x^{7}-y^{7}.$
$\Rightarrow$ We have that $p/x+y$ or $p/xy$ or $p/x^{2}+xy+y^{2}$ then it is obvious.
$\Leftarrow$ We have that $p/x+y$ or $p/xy$ or $p/(x^{2}+xy+y^{2})^{2}.$ For the first two cases it is obvious that the statement is true. For the last one it follows from tha fact that if $p$ is a prime and $p/a^{2}$ then $p/a.$
Thus 29 is nice and there exist infinitely many nice numbers(every prime $>7$).
Since $2006=2.17.59$ it is similar.
But 2007 is not nice :P


function
factorial
algebra
polynomial
set theory
irrational number

I've been reading about set theory, cardinalities inifinte sets, etc. I have a few questions beyond the scope of the books i've been looking through.

1. What is the cardinality of the set of digits following the decimal point in an irrational number (eg, 3.1415...)? I suspect it's aleph-null (denumerable). But is it possible to have other cardinalities for this set?
2. There are infinite sets that are not denumerable (such the set of reals, cardinality c). In other words, there are infinite sets that cannot be put into one-to-one correspondence with the natural numbers. But are there 'uncountable' [i]finite[/i] sets, ie finite sets that cannot be put into one-to-one correspondence with any finite subset of the natural numbers? If so, what is an example?
3. on the complex number plane, where do the cardinalities of infinite sets (eg, aleph-null, c) lie (if anywhere)? are these 'transfinite' numbers real, complex, niether?
4. What is the cardinality of the infinite set of all transfinite numbers (or is that linked to acceptance/rejectance of continuum hypothesis axion?)?
5. What happens when you add a finite number to a transfinite one?

Any information is welcome. Thanks.

I moved this thread from the Round Table to here, where it's more appropriate.

My answers will make reference to the ordinals.  You really should know about how the ordinals work before you start trying to figure out cardinals.

1) Yes, any list is countable.  The digits following the decimal point are a list, thus countable in number.  If you wanted "longer" decimals, you'd have to figure out what they meant.  In particular, this would require you to figure out what it means to raise 2 (or 10, or other such things) to the power of a non-finite ordinal.  Conventionally, the result won't have a multiplicative inverse, and so the result won't have meaning.  In the surreal numbers, there are such numbers, I believe.

2) No.

3) This question doesn't make sense to me.  [i]Every[/i] point in the complex plane represents a finite complex number.

4) It isn't: the class of transfinite numbers (ordinals, etc.) is larger than set-size.  Since it's not set-size, it doesn't have cardinality.

5) It depends what order you add them in -- ordinal arithmetic is non-commutative.  $\omega + 1$ is the successor ordinal to $\omega$, so different from and larger than $\omega$, but $1 + \omega = \omega$.

Here are three more questions on Cardinal Numbers that I've been thinking about.

1)
What is $\aleph_1!$. I suspect that it is $\aleph_2$, but I have no reason to believe this.

2)
All positive integers can be represented in binary putting a string of $\aleph_0$ zeros on one end so that they are all the same length. There are by definition $\aleph_0$ integers. However, each of the $\aleph_0$ spaces for binary digits can either be a 0 or a 1, giving the possibility of $2^{\aleph_0}=\aleph_1$ integers. How does that work?

3)
We know that there are $\aleph_0$ primes because they can be put in a one-to-one correspondence with the integers. If we take the power set of the set of primes (ie, the set of all subsets of primes), we have a set of order $2^{\aleph_0}=\aleph_1$. But this has the same number of elements as the square-free numberss (if each set of subsets of primes is written as the product of the primes in that set). And we know that the order of the square-free numbers is less than or equal to the order of the integers. Does this mean that $\aleph_1<\aleph_0$?

1.  To clarify: when you write $\aleph_1$, you mean the first uncountable ordinal, yes?  How are you defining the function $!$?

2. Most of those strings have 1s in arbitrarily high positions -- those strings do not represent integers.  If you restrict yourself to strings where eventually all terms are zero, then you're back with things that represent integers, and also back with countably many of them.

3. Same problem here: you've included subsets which contain infinitely many primes, and these subsets do not represent integers.

$\aleph_1$ is $2^{aleph_0}$. $\aleph_0$ is the order of the set of positive integers. $\aleph_1$ is also the order of the set of reals.

! is the factorial function, but i'm not sure if i want to define it in ascending or descending order. in fact, i dont even know how either of those affect the problem. assume whichever way makes it easier to solve.

[quote="gopherhole112"]$\aleph_1$ is $2^{\aleph_0}$. $\aleph_0$ is the order of the set of positive integers. $\aleph_1$ is also the order of the set of reals.

! is the factorial function, but i'm not sure if i want to define it in ascending or descending order. in fact, i dont even know how either of those affect the problem. assume whichever way makes it easier to solve.[/quote]

Your second sentence follows from your first only if we take the Continuum hypothesis to be true -- watch out for that.
More importantly, I don't really think this is the definition you want to use.  It would probably be much safer to define things in terms of ordinals, not cardinals, since the rules of arithmetic make more sense there (and since it's presumably the ordinals which you want to be multiplying together, not the cardinals).  Then we could have $\aleph_0 = \omega$, the first infinite ordinal, and $\aleph_1$ is $\omega_1$, the first uncountable ordinal.

Also, I think for a variety of reasons I don't want to use the standard definition of $!$ here and instead use something like $(n - 1)!$: for an ordinal $\alpha$, we want to consider the $p(\alpha)$, the product of all positive ordinals strictly less than $\alpha$, where for limit ordinals $\lambda$ we take $p(\lambda)$ as the least upper bound of the set $\{p(\beta)\mid \beta < \lambda\}$.  Thus $p(3) = 2$, $p(5) = 24$ and $p(\omega) = \omega$.  (Otherwise, I wouldn't have a convenient expression for "the product of all ordinals less than a given limit ordinal.")

Now, in computing $p$, it does matter which order we multiply ordinals in, after we get to $\omega + 1$.  In particular, $p(\omega) = \omega, p(\omega + 1) = \omega^2$ but $p(\omega + 2)$ could be either $\omega^2\cdot(\omega + 1) = \omega^3 + \omega^2$ or $(\omega + 1)\cdot\omega^2= \omega^3 + 1$ which are not equal to each other.  This difference will presumably persist for larger ordinals.  However, it isn't likely to make a substantial difference in the order of the values of $p$ on large ordinals -- I would expect the difference between the two to be relatively small compared to the value of either.  (It is possible that I'm totally wrong about this, of course.)  Now, coming to your question: I don't have the faintest idea what the value will be, but my guess is that it will be vastly smaller than what you expect.  In particular, if $\omega_1$ is the first uncountable ordinal, $p(\omega_1)$ is the product of $\omega_1$-many ordinals, and each of them is only countable -- I wouldn't expect that product to get us all the way to a new order of infinity (and I certainly wouldn't expect it to come out to some nice-looking ordinal like $\omega_2$).

One other important thing: in ordinal arithmetic, $2^\omega$ is not the same thing that we mean when we write $2^{\aleph_0}$.  Exponentiation for ordinals is just an operation defined by transfinite recursion: if $\alpha$ is a positive ordinal then
$\alpha^0 = 1$
$\alpha^{\beta + 1} = \alpha^\beta\cdot\alpha$ and
$\alpha^\lambda = \textrm{l.u.b.}(\{\alpha^\beta\mid \beta < \lambda\})$ where l.u.b. means the least upper bound of that set.  In particular, for this operation we have that $2^\omega = \textrm{l.u.b}(\{1, 2, 4, 8, \ldots\} = \omega$, not $\omega_1$.

wow, you and i learned it a bit differently. i was able to follow what you said, but (believe it or not) i learned it solely with $\aleph$s. No $\omega$s. I was also under the assumption that the continuum hypothesis had been proven true (by godel i think), but i could be wrong.

It is quite reasonable to learn it with just cardinals -- I just happen to be taking two classes right now that use the ordinals fairly extensively and (having now done it both ways) it makes much more sense to me to do it with ordinals and then define cardinals in terms of ordinals (a cardinal is the smallest ordinal which isn't in bijection with any smaller ordinal).  If we do it your way, though, I'm not sure what meaning to give to the factorial expression: there aren't enough cardinals to make $\aleph_1!$ very interesting, I think, since the product of the countably many finite cardinals is going to be countable, then you multiply by $\aleph_0$ which keeps you countable, and then you multiply by $\aleph_1$, which gets you to order $\aleph_1$.  The last term will always be vastly larger than the product of all preceding terms.  (Maybe you had something else in mind -- if so, I'd be happy to hear what it was.)

Godel proved that the continuum hypothesis was [i]consistent[/i] with the other axioms of set theory -- if one assumes it true, there aren't any logical contradictions that arise.  However, Paul Cohen proved that its negation was also consistent -- that is, if one assumes the CH is false, no logical contradictions arise, either.  Thus, it's independent of the other axioms of (Zermelo-Fraenkel) set theory.  Unlike the Axiom of Choice, which also deals with infinite sets and is independent of the axioms, the CH has not been accepted into the canonical set of axioms of set theory.  (Most set theory is done in the ZFC system -- Zermelo-Fraenkel plus Choice.)  In some models the CH is true, in others it is false, and in plenty it is undecideable.  In fact, it seems that (for whatever reason) set theorists seem to have reached a partial consensus that the CH should be regarded as false.  I don't know why this is the case, and set theorists are a rather odd bunch, but there you have it.

Here was my idea for finding $\aleph_1!$: Cartesian cross products.

Let $a_i$ refer to a set with exactly $i$ elements.
And let $b$ be defined by:
$b\equiv a_1\times a_2\times a_3\times\cdot\times a_{aleph_1}$
(I did a little bit of telescoping there).
So The order of $b$ is $\aleph_1$, but I haven't been able to get anywhere with this method.

*Just to make sure, the cross product of sets $a$ and $b$ is defined as $a\times b\equiv{(x,y)\mid x\in a}$ and ${y\in b}$

It might also help for me to explain where I came up with this problem. A friend of mine pointed out that $\aleph_0$ is generally associated with the integers. $\aleph_1$ is generally associated with the reals. But there is no set that $\aleph_2$ is associated with. (It's the power set of the reals, but he wanted a something more concrete.) So the quest was to find a 'concrete' set of order $\aleph_2$. My idea was the set of all polynomials with real coefficients in one variable. Here's what I've found so far:

First I looked at it by varying the coefficients. For each power of $x$, we had a  choice  of $\aleph_1$ reals for the coefficients, and there are $\aleph_0$ powers of $x$, so we have $\aleph_0\cdot\aleph_1=\aleph_1$. It seems like we've proven that there are $\aleph_1$, but lets do it another way.

There are $\aleph_1$ different possible roots, each of which can have a multiplicity of somehwere between 0 and $\aleph_0$... which gets us $\aleph_1$ again.

Now I'm confused. somehow I got the term $\aleph_1!$

Must have done something wrong. oh well, I still want to know what $\aleph_1!$ is, and I still want a concrete example of $\aleph_2$ if you can find one.

Okay.  So, as your analysis suggests, if we take the cartesian product of sets of order $i$ for integers $i$, and a set of order $\aleph_0$ and a set of order $\aleph_1$, we spit out a set of order $\aleph_1$.  This is because the first part of this product (the finite sets) just gives us something of order $\aleph_0$ (not very big) which we then cross with another countable set (so we're still countable) and then finally cross with something of order $\aleph_1$.  Because $\aleph_1$ is so huge in comparison with $\aleph_0$, the result of this is a set of order $\aleph_1$ (the other parts of the product don't matter at all).  And this generalizes to $\aleph_i$ for each integer $i$ -- $\aleph_i! = \aleph_i$ (using your definition of the $!$ function).

$\aleph_2$ is really, really, really big.  But, remember that the meaningfulness of your question (for a concrete example) relies upon accepting the continuum hypothesis -- if we take the CH to be false (as most set theorists seem to do, when it matters), then $\bf R$ is actually an example of a set of order $\aleph_2$, or perhaps even larger.


Facebook/email of Indiana ARML

anybody know if it exists?

No...well I dont have a facebook, but Im pretty sure we never had a group or something like that.

Just keep this thread alive.  I'll try to post more information sometime in early April.

Consider this our group lol.


Em đang cần lời giải những bài sau,được bài nào hay bài ấy
 bài 1
tìm tất cả các hàm số f xác định và khả vi trên [0,1] thỏa mãn
 a) f(0)=f(1)=1
 b) 2003f'(x)+2004f(x)$ \geq$2004    $ \forall$x $ \in$[0,1]

 bài 2
giả sử hàm số f(x) khả vi trên [a,b] thỏa mãn
     [f(x)]^{2} + [f'(x)]^{2}>0 với mọi x $ \in$[a,b]
 chứng minh rằng số nghiệm của phương trình f(x)=0 là hữu hạn
bài 3
cho a thuộc(0,1).hàm số f(x) liên tục trên [0,1] thỏa mãn f(0)=f(1)=0
chứng minh rằng  $ \exists$ b $ \in (0,1)$ sao cho f(b)=(f(b-a) hoặc f(b)=f(b+a-1)
 bài 4
cho f(x) khả vi vô hạn trên R thỏa mãn
a)tồn tại M>0 sao cho -M<  f{n}(x)<M vói mọi x $ \in$ R và n $ \in$N
b)$ f( \frac{1}{n})\equal{}0$ với mọi n $ \in$N
chứng minh rằng f(x)=0 với mọi x

 trên diễn đàn toán học chả có ai chịu làm cả

Người ta không chịu làm vì đây toàn là các bài Olympic sinh viên mấy năm trước, em vào box tương ứng trên Mathlinks nhé!

cảm ơn anh nhiều,thảo nào mấy bài này em không tìm được trong sách phổ thông,anh có thể dẫn link cho em được không

Đây này http://www.mathlinks.ro/Forum/index.php?f=81 .  :lol:

[hide="Bài 1"]1) xét $ g(x) \equal{} e^{cx}f(x)$ trong đó $ c \equal{} \frac {2004}{2003}$ thì $ g'(x)\geq ce^{cx}$ với $ x\in[0;1]$. Lấy tích phân từ $ 0$ đến $ x$ và từ $ x$ đến $ 1$ được kết quả.[/hide]

[hide="Bài 2"]2) Hình như bài này cần giả thiết $ f$ là lớp $ C^1$? Nếu số không điểm của $ f$ là vô hạn thì dãy không điểm này có 1 điểm tụ, có nghĩa có thể lấy ra 1 dãy con vô hạn hội tụ. Sử dụng định lý Rolle trên đoạn $ [a_n,a_{n \plus{} 1}]$ của dãy con này sẽ dẫn tới tồn tại $ x_0$ để $ f(x_0) \equal{} f'(x_0) \equal{} 0$.[/hide]

mình đã search link nhưng chỉ tìm được gợi ý cho bài 2,các dữ kiện đều chính xác,ta có thể chọn ra dãy x(n) hội tụ trong dãy các nghiệm của f(x) nếu f(x) có vô số nghiệm.Cảm ơn lời gợi ý của
pluricomplex

các anh làm giúp em thêm mấy bài này
bài 1
giả sử f(x) liên tục trên[0,1] và có đạo hàm trên (0,1);thỏa mãn f(0)=0
chứng minh rằng  $ \int\limits_{0}^{1}/f(x)f'(x)/d(x) \leq \frac {1}{2} \int\limits_{0}^{1}( f'(x)^{2}d(x)$
bài 2
giả sử f(x) xác định và có đạo hàm cấp 2 trên R thỏa mãn $ f(x) \plus{} f''(x) \geq 0 \forall x$chứng minh rằng
 $ f(x) \plus{} f(x \plus{} \pi ) \geq 0 \forall x$
bài 3
cho hàm số f(x) liên tục trên [0,1] và có đạo hàm bậc 2 trên khoảng này,thỏa mãn $ f(0)\equal{}f(1)\equal{}0$
$ minf(x)\equal{}\minus{}1$.Chứng minh rằng $ max /f''(x)/ \geq 8$


modular arithmetic
number theory proposed
number theory

Find all n,m,k>0 such that \[5^n-3^m = k^2\]
Although it was proposed for students between 15 and 16 years old, nobody could solve it. Let see if u can  :lol:

Rewriting it as $5^n=3^m+k^2$ it's easy to see solution: n=2 m=2 k=4. To prove that there is no other solution I guess we should look at this equation: mod 3, mod 4 and mod 5.

I am solving $5^m=k^2+3^n$.

1. Using $\pmod{3}$ we obtain that $m$ is even.
2. Using $\pmod{4}$ we obtain $n$ is even.

So $m=2m'$, $n=2n'$ and $(k,3^{n'},5^{m'})$ is a Pythagorean triple. It follows that $5^{m'}=a^2+b^2$ and $3^{n'}=a^2-b^2$ OR $3^{n'}=2ab$, though second case is inadmissible.
Therefore, we see that $a+b=3^x$, $a-b=3^y$ and $2\cdot 5^{m'}=2(a^2+b^2)=3^{2x}+3^{2y}$, thus it is necessary $y=0$. Hence we obtain equation $3^{2x}+1=2\cdot 5^{m'}$.
Suppose $m'>1$ is a minimal solution of mentioned equation and consider $\pmod{25}$. It yields that $5\,|\,x$, so $x=5l$ and $3^{2x}+1=(3^{2l}+1)(...)=2\cdot 5^{m'}$. It is necessary $3^{2l}+1=2\cdot 5^{m''}$, $m''<m'$. Thus it must be $m''=1$, and then $l=1$ $\Rightarrow$ $x=5$, but $3^{10}+1\neq 2\cdot 5^{m'}$.

Therefore, unique solution of $3^{2x}+1=2\cdot 5^{m'}$ is $x=1$ and $m'=1$. From here we obtain $a=2$ and $b=1$. Finally, unique solution of the initial equation is $m=n=2$ and $k=4$.

Thanks Myth - this is exactly what I thought of  :D

Only remark: I am far from 15-16 years old  ;)  :D  :D


Find all function $f: R^{+}\rightarrow R$ satified condition:
$f(x+y)+f(\sqrt{xy})=f(x^{2}y)+f(xy^{2})$
$R^{+}$:set of positive reals

Mmm... I'm stuck with a bunch of unrelated facts:

$y=x$ $\to$ $f(2x)+f(x)=2f(x^{3})$.
$y=1$ $\to$ $f(x+1)+f(\sqrt{x})=f(x)+f(x^{2})$.
...
with the special cases
$y=x=1$ $\to$ $f(2)=f(1)$.
$y=x=\sqrt{2}$ $\to$ $f(2\sqrt{2})=f(\sqrt{2})$.
$y=1$, $x=\phi$ $\to$ $f(\phi)=f(\sqrt{\phi})$, where $\phi$ is the golden ratio.
...

Since $1<\sqrt{\phi}<\sqrt{2}<\phi<2<2\sqrt{2}$,
it makes me think there is no "easy" solution, other than the trivial
$f(x)=k$, for any $k\in\mathbb{R}$.
But, of course, I'm still far from the proof.

Can anyone help us to solve this problem ?:maybe:

Here's another fact (not very useful, but it makes me think more and more that there is no other "easy" solution that isn't constant):
Every value in the range of $f$ (except at most three) is reached at least four times.
[hide="Proof"]$x=z^{3/5}$, $y=z^{-1/5}$ $\to$ $f(z^{3/5}+z^{-1/5})=f(z)$.

If we define $a(z)=z^{3/5}+z^{-1/5}$, we have $f(z)=f(a(z))$.

It can be seen that there's only one point with $a(z_{p})=z_{p}$.

Also, $a(z)$ has a minimum at $z_{m}$, and for every $z\neq z_{m}$ , there is one $z'\neq z$ with $a(z')=a(z)$.

Therefore, for every value $z_{0}$ (except those related to $z_{p}$ or $z_{m}$),
$f(z_{0})=f(a(z_{0}))=f(a(z'_{0}))=f(z'_{0})$[/hide]

$f(x+y)=f(x^{2}y)+f(xy^{2})-f(\sqrt{xy})$.

The only continuous solution at every point,
if the limit $L=\lim_{t\to 0^{+}}{f(t)}$ exists and is finite,
must be, for every $x>0$,
$f(x)=\lim_{y\to 0}{f(x+y)}=\lim_{y\to 0}{(f(x^{2}y)+f(xy^{2})-f(\sqrt{xy}))}=L$.
With these restrictions, $f$ can only be constant.
(Of course, for now it's only a special case.)

Does The probem has to add contionuos condition? I think so, but if add more condition then the problem really is very dificult


analytic geometry
geometry
3D geometry
conics
ellipse
linear algebra

¿which points X satisfy r=0 where (r,0) are the polar coordintes of X? 0= is supposed to be the greek letter theta. sorry for the inconvenience.

 also if were working in cylindrical coordinates in space ¿what would be the graph of z=r. here X= (x,y,z)= (r,0,z)

 where (r,0) are the polar coordinates of point projection of X onto the plane xy and z=z. again the 0 is supposed to be the greek letter theta.

It's Archimedes' spiral: $ r \equal{} a\theta$. (special case of it: $ a \equal{} 1$).

[url]http://mathworld.wolfram.com/ArchimedesSpiral.html[/url]

if r=z we get z^2 = r^2 = x^2 + y^2 a paraboloid

Isn't $ z \equal{} x^2 \plus{} y^2$ a paraboloid, and you here have $ z^2 \equal{} x^2 \plus{} y^2$. It is an elliptic cone (to be more precise, it's ellipse is a unit circle).

[url]http://mathworld.wolfram.com/EllipticCone.html[/url]


A bug starts at $(0,0)$. It can make three moves: a step right of length 1, a step left of length 1, or stay put. He is twice as likely to stay put as to go left, and he is just as likely to go left as right. Compute the probability that the bug lands on $(i,0)$ after $n$ moves.

[hide] Obviously, the answer is the coefficient of $x_{i}$ in $(\frac{1}{4x}+\frac{1}{2}+\frac{x}{4})^{n}$. Is there a way to calculate this coefficient besides multiplying out? [/hide]


Let $A$ be the set of all rational numbers greater than 1.

Can $A$ be partitioned into two sets $X,Y$ such that the sum of any two numbers from one subset belongs to the same subset?

If such a partition does exist then $X$ and $Y$ are stable by finite sums.

Assume that $2 \in X$, then $2p \in X$ for all positive integer $p$.
Now, assume that $\frac{a}{b}\in Y$, then $2a = 2b\times \frac{a}{b}\in Y$, which contradicts that $X \cap Y = \emptyset.$

Pierre.

[quote="Jumbler"]Let $A$ be the set of all rational numbers greater than 1.
[/quote]

I think you miss something Rust  :P

Pierre.

It is obviosly if $A=\{x\in Q|x>0\}$ and $r\in X$, then A=X.
If A is Q, then we have only 2 cases $X=\{r>0\}, Y=\{r\le 0\}$ and $X=\{r\ge 0\}, Y=\{r<0\}$.

What if "sum" is changed to "product"?  :)


Let P be a point in the interior of convex quadrilateral ABCD. Prove that $\text{Min}\{\angle PAB, \angle PBC, \angle PCD, \angle PDA\}\leq \frac{\pi}{4}$

I believe this has been solved here: http://www.mathlinks.ro/Forum/viewtopic.php?p=17364

  darij


calculus
integration
calculus computations

Show how to approximate the required work by a Riemann sum. Then express the work as an integral and evaluate it.

A heavy rope, 50 ft long, weighs .5 lb/ft and hangs over the top of the building 120 ft high.

(a). How much work is done in pulling the rope to the top of the builing?

(b). How much work is done in pulling half the rope to the top of the building?

Here is what I did.

Fk=120ft * 25 lbs= 3,000 ft-lb

dk=.5 $120\int50$ (120-y)

120y- $\frac{y^{2}}{2}$

then i plugged in 120 into the y's and i got 7,200 * .5=3600

W total= 3,600 + 3,000=4,600 ft-lb

but the real answer is 625 ft-lb,

what did i do wrong and can you show me how to do it

Look closely at your integral: what are the limits? That's where your problem is.

No, it wouldn't- or if it is, you should be integrating a different function. What exactly is $y$ in your integral?

jmerry is trying to help, he also is very good at math, so he's asking you question to guide your thought process so in a way you figure it out


complex analysis
function
algebra
domain
functional equation
number theory theorems
number theory

$\zeta(x)=\sum_{n=1}^{\infty}\frac{1}{n^{x}}$
that is true only for$Re(x)\gneq1$ :lol:

Hmm, I think I've to answer here :D

For $\mathfrak{Re}(x)>1$, we have \[\zeta(x)=\sum_{n=1}^{\infty}\frac{1}{n^{x}}\] It is a holomorphic function, meaning that it is complex differentiable in it's whole domain (such functions have a lot of nice properties, e.g. they are $\infty$ often differentiable).
Like most holomorphic functions, it can be extended (without loosing properties) to a bigger domain than the initial one. In fact, it can be extended to the whole complex plane.

There is a functional equation which looks, if written in terms of $\zeta$, rather ugly. But with $\Psi(x) : = \zeta(x) \Gamma\left( \frac{x}{2}\right) \pi^{-x}$, it gets: \[\Psi(x)=\Psi(1-x)\] It is e.g. used in most proofs of the prime number theorem, but also occcurs in proofs of Dirichlet's theorem on primes in arithmetic progressions.
There are some generalisations of such functions, e.g. the L-series or Dedekind Zeta functions of number fields.

But taking a look at the follwong links (the first one is best) would be better than I could explain:

http://mathworld.wolfram.com/RiemannZetaFunction.html
http://mathworld.wolfram.com/DirichletL-Series.html
http://en.wikipedia.org/wiki/Zeta_function


A similar problem was discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=factorial+equation+a%21%3Db%21%2Bc%21&t=50310]here.[/url]


what are the condition necessary for nucleophilic substitution of allyl alcohol in presence f $ HBr$ than electrophillic addition?

From my point of view, nucleophilic substitution is already favoured, because the oxygen atom protonation is preferred over double bond protonation. However, for a general allylic alcohol, we should always expect some allylic rearrangement to take place.


calculus
derivative
quadratics
limit
geometry
geometric transformation
symmetry

Let the roots of the equation $x^3+ax^2+bx+c=0$ are all real.
Show that the difference between the greatest and the least roots lies between
$\sqrt {a^2  - 3b}$ and $2\sqrt {(a^2  - 3b)/3}$.

I tried forming an equation whose roots are the differences of the roots of the given equation, but could not arrive at the solution.

Since the cubic has 3 roots, it must have 1 local minimum and one local maximum. These are found by putting the derivative to zero:

$\frac{\text dy}{\text dx} = 3x^2 + 2ax + b = 0$

$m_{12} = \frac{-2a \pm \sqrt{4a^2 - 12b}}{6} = \frac{-a \pm \sqrt{a^2 - 3b}}{3}$

The roots $m_1 < m_2$ of this quadratic equation are the positions of the minimum and maximum. Since the coefficient of the cubic at the $x^3$ term is positive, $\lim_{x \rightarrow +\infty} y(x) = +\infty$ and $\lim_{x \rightarrow -\infty} y(x) = -\infty$. Clearly, the smaller root $m_1$ is the position of the maximum, while the larger root $m_2$ is the position of the minimum. The difference between the minimum and maximum positions is

$\triangle m = m_2 - m_1 = \frac 2 3 \sqrt{a^2 - 3b}$

The graph of the cubic y(x) is symmetrical with respect to the point

$M = \left[\frac{m_1 + m_2}{2},\ y\left(\frac{m_1 + m_2}{2}\right)\right]$

and it is useful to put this point on the y-axis by translating the graph by $\frac{m_1 + m_2}{2} = -\frac a 3$:

$u = x - \frac{m_1 + m_2}{2} = x + \frac a 3,\ \ x = u - \frac a 3$

Substituting this to the cubic,

$\left(u - \frac a 3\right)^3 + a \left(u - \frac a 3\right)^2 + b \left(u - \frac a 3\right) + c = 0$

$u^3 - au^2 + \frac{a^2}{3} u - \frac{a^3}{27} + au^2 - \frac{2a^2}{3} u + \frac{a^3}{9} + bu - \frac{ab}{3} + c = 0$

$u^3 + \left(b - \frac{a^2}{3}\right) u + c - \frac{ab}{3} + \frac{2a^3}{27} = 0$

The quadratic term of the cubic disappeared. Since neither the difference $\triangle m$ between the minimum and maximum positions, nor the difference $\triangle r$ between the maximum and minimum roots of the cubic changed in this translation, we can just go ahead and investigate the cubic

$y = x^3 + \frac{3b - a^2}{3} x + d$ [color=white].[/color] where [color=white].[/color] $d = c - \frac{ab}{3} + \frac{2a^3}{27}$

The x-axis cuts this cubic somewhere between the horizontal tangents to the graph at the maximum and minimum. If the translated point M falls on the x-axis, i.e., into the origin, then the the cubic passes through the origin, which means that d = 0, the middle root $r_2 = 0$ and the minimum and maximum roots $r_1, r_3$ are the same in absolute value and have opposite signs, because of symmetry WRT the point M, the origin, $r_1 = -r_3$. So, now we have the cubic eqution for the

$y = x^3 + \frac{3b - a^2}{3} x = x \left(x^2 - \frac{a^2 - 3b}{3}\right) = 0$

with the roots $r_3 = \sqrt{\frac{a^2 - 3b}{3}}> 0,\ r_2 = 0,\ r_1 = -r_3 < 0$. The difference between the minimum and maximum roots is

$\triangle r_{\text{max}} = r_3 - r_1 = 2r_3 = 2\sqrt{\frac{a^2 - 3b}{3}}$

This clearly is the maximum possible difference between the roots (see below). Now we move the graph of the cubic up or down WRT the x-axis (i.e., change the y-intercept d), while the x-axis stays between the horizontal tangents at the maximum and minimum (i.e., the cubic still has 3 real roots). Because of the symmetry of the graph WRT the point M, we can look just at moving the graph down (let d < 0), until the x-axis becomes the horizontal tangent at the maximum $m_1 < 0$. In this process, we cut off a portion of the cubic left of $m_1$ and added a portion of the cubic right of $m_2$ to the difference between the new roots $r_1, r_3$. Since the cubic grows faster in the added region than it drops in the cut-off region, the new roots $r_1, r_3$ move closer together, $\triangle r  = r_3 - r_1 < \triangle r_{\text{max}}$. When the x-axis becomes the horizontal tangent to the graph at the maximum, we reached the minimum difference between the roots and $r_1 = r_2 = m_1$ becomes a double root of the cubic. The position of the maximum is

$m_1 = -\frac{\sqrt{a^2 - 3b}}{3}$

(either subtract $-\frac a 3$ (i.e., add $+\frac a 3$) to the position of the maximum before the translation by $-\frac a 3$ or differentiate again). The value at the maximum, when the x-axis passed through the point M, was

$y(m_1) = -\frac{\sqrt{a^2 - 3b}}{3} \left(\frac{a^2 - 3b}{9} - \frac{a^2 - 3b}{3}\right) = \frac{2 (\sqrt{a^2 - 3b})^3}{27}$

Hence, now $d = -\frac{2 (\sqrt{a^2 - 3b})^3}{27}$ and the cubic equation

$y = x^3 - \frac{a^2 - 3b}{3} x - \frac{2 (\sqrt{a^2 - 3b})^3}{27} = 0$

has a double root $r_1 = r_2 = -\frac{\sqrt{a^2 - 3b}}{3}$. Substituting $\xi = \frac{3x}{\sqrt{a^2 - 3b}}$, the cubic equation becomes

$\xi^3 - 3 \xi - 2 = 0$

with a double root $\xi_1 = \xi_2 = -1$. Factoring it out by long division,

$\xi^3 - 3 \xi - 2 = (\xi - 2) (\xi^2 + 2 \xi + 1) = 0$

Thus the remaining root of our cubic equation, when the x-axis is tangent to the maximum, is $r_3 = \frac{\sqrt{a^2 - 3b}}{3} \xi_3 = \frac{2 \sqrt{a^2 - 3b}}{3}$ and the minimum difference of the roots is

$\triangle r_{\text{min}} = r_3 - r_1 = \frac{2 \sqrt{a^2 - 3b}}{3} + \frac{\sqrt{a^2 - 3b}}{3} = \sqrt{a^2 - 3b}$

Great solution Yetti.  I really enjoyed going through it.

Thanks


AMC
AMC 10
AMC 12
ARML
trigonometry
AMC 12 A
MATHCOUNTS

Would have been 10A state winner in Georgia, but I moved to Texas. Oh well.

Rawr, first on 12A/B. Our 12B team narrowly edged out our 12A team for the CT title.

haha, (^_^) and I beasted AMC 12B in our state

the 8th grade 12B average ownz all other grade averages for 12B
not that I had anything to do with it...

There seem to be quite a few errors...I found the number of people under grades and the number of people under genders in like more than 4 places. And I only skimmed the MA page.

[quote="tenniskidperson3"]Top 10B score in Ohio.[/quote]

Lol 2nd 10B score in Ohio. Dang you 2^4

Oh dang like 1337th place for 10A. roflcopter fail

[quote="Yongyi781"]Hm apparently I am the only 9th grader from MA who got 150 on the AMC 10A.[/quote]

lol me and someone else in my school (both 9th graders) both got perfect scores on the 10A officially. I wonder what this means.

Wow I'm the one ninth grader to have gotten a 115.5 on the 12.

EDIT: Doh wow this is NJ. I fail at life.

Mistakes like these always make me fail. I don't know how I got a perfect on the AMC. lol.

[quote="E^(pi*i)=-1"]Why does Maryland always have such low AMC scores?  Double 129's are good enough to get 3rd on the 12A and tie for 2nd on the 12B?  :huh:[/quote]
 :lol: did you get the 150?

[quote="HiDN428"]How do u guys know how many people per took amc per country....i mean like global statistics :?:  :help:[/quote]

Go to [url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2009-12a/09-1-1012-StateStats.shtml]AMC State Statistics Page[/url] and scroll down; you will see global statistics available under "International Schools".

[quote="ProtestanT"]Looking at these stats, I realize that barely 1% of the High School population in the US take the AMCs...and this is the pool used to select the USAMO qualifiers and ultimately the team for IMO... If more of the other 99% are told about these contests, maybe we can have a lot more good people trying these tests... It is very likely there are more good math people in the other 99%.[/quote]

From what I know from our district high school (one of the largest in our state), only the math team and math clubs members participate in the AMC contests. Though the number is small, these students do represent the best students in our district high school. For example, this year, we have more than 30 AIME qualifiers, and almost every National Merit Scholarship recipient (we have 25 this year) is AIME qualifier. In our district high school, students also participate in many regional, state and national math contests, and pretty much all the top students are also the AIME qualifiers.

If you take a look at the other state and national level math contests, such as HMMT, Mandelbrot, USAMTS etc. there is very exception that the top students are not part of the AMC/AIME/USAMO contests.

In my opinion, Dr. Dunbar and his AMC organization has done a tremendous amount of good job keep the AMC/AIME/USAMO contest going every year and they do deserve our respect.

[quote="shelly32494"][quote="Textangle"]

EDIT: WOAH look at CHINA...[/quote]

omg. holy crap :xx
thats INSANE.[/quote]

My father was born in China and got that sort of training. Though he has not actively done any math competition since graduate school days 20+ years ago, he can still beat me easy on Trig and Geometry. He keeps telling me, young boy, someday when you can defeat me on Geometry and Trig, you will surely make into USAMO easily :-). One thing I did notice of him is he never used calculator, and whatever he knows how to do it, he will never make any mistakes on it.

I guess that is something we ought to learn from China :-). Good luck to all!!!!!!

highest eighth grader in MI for both i took (by quite a bit), highest overall on 10A (144), somewhere around sixth for 12B (121.5). yay?

me and a couple billion others in texas got a 150 on the 10a. i wont elaborate on the 12b  :rotfl:

[quote="soltixnj"][quote="ronblumen"]

If you take a look at the other state and national level math contests, such as HMMT, Mandelbrot, USAMTS etc. there is very exception that the top students are not part of the AMC/AIME/USAMO contests.

[/quote]

Do you have any proof of this? I really don't think that this is true.[/quote]
Yes, it is true.

For 2008-2009 Mandelbrot national level contest (http://mandelbrot.org/), among the top 10 winners, at least 8 are USAMO qualifiers.

For USAMTS, the only three perfect scorers (http://www.usamts.org/Leaders/U_Leaders.php) are all USAMO qualifiers.

For 2009 HMMT (http://web.mit.edu/hmmt/www/datafiles/results/2009ext.shtml), since two best high schools in China (Fudan Fu Zhong & Shanghai High School) competed this year, some of the top winners can not qualify for USAMO. But if you exclude those foreign students, plus a few IMO students who just migrated to Phillip Exeter this year, most of the top USA winners in each of the categories also qualified for USAMO.

Noting participation in other contests still doesn't address the issue of the students who are talented but don't do [b]anything[/b]. Countries like Bulgaria are far better at finding talent, as they do much better at IMO relative to population size.

@worthawholebean: i don't think imo participation relative to population is a very good measure of finding talent. all that says is that bulgaria has amazing support for its top tier of math students. imo is 6 people- better talent finding in the u.s. won't drastically increase our scores. however, i do think the u.s. could improve its education system and economic competitiveness by finding talent better through math competitions. how to do that w/o forcing people, i have no idea...

[quote="Yongyi781"]What I find particularly funny is that 4 people, overall, actually managed to get a score of 0. Statistically, the probability that one would get a 0 by randomly guessing is $ \left(\frac 45\right)^{25}\approx 0.0038 \approx \frac 1{265}$ chance.

Also, exactly one student from Russia took the AMC's at all. You can find out what answer choices he picked based on the answer choice distribution :D[/quote]

If I counted correctly I believe only 3 people got zeroes.

Hm. 10A winner.
Not that it means much now...

It is true there were 203 perfect scores on AMC 10A/B.  But I found a lot of them were international students.  For example, there were 66 perfect scores on 10B from China.

[quote="IDEALLY.PERFECT"]200+ perfect scorers on amc 10A/B
...was definitely an easy year for amc 10[/quote]

In America, there were 49 perfect scorers on the 10A and 44 perfect scorers on the 10B.


abstract algebra
group theory
superior algebra
superior algebra unsolved

If $ p <  q$ are primes, prove that a group of order $ pq$ must be cyclic unless $ p \mid (q \minus{} 1)$. Construct
a non-abelian group of order $ pq$ when $ p \mid (q \minus{}1)$

[quote="karunya"]If $ p < q$ are primes, prove that a group of order $ pq$ must be cyclic unless $ p \mid (q \minus{} 1)$. Construct
a non-abelian group of order $ pq$ when $ p \mid (q \minus{} 1)$[/quote]

non-abelian group... the dihedral group is an example with $ p\equal{}2$.

Put $ a^q\equal{}1$, $ b^p\equal{}1$, $ bab^{\minus{}1}\equal{}a^i$ where $ i^p\equiv1(\mod q)$, $ i\neq1(\mod q)$.

i don't get it. can u state all the results you are using alongside. sorry, but i'm really new to group theory :blush:  :oops:

I use: sylow theorems, chinese remainder theorem, semidirect products (internal characterization and universal property), presentations (universal property)

if $ p$ does not divide $ q\minus{}1$, show that there are exactly one $ p$- and $ q$-sylow subgroup. then $ G$ is the direct product of them and thus cyclic.
if $ p$ divides $ q\minus{}1$, then $ G$ possibly also has unique sylow subgroups and is cyclic, or there is exactly one $ q$-sylow subgroup $ N$ and exactly $ q$ $ p$-sylow subgroups $ U$. then $ G$ is the semidirect product of $ N$ with $ U$, which is characterized by a nontrivial homomorphism $ U \to Aut(N)$, i.e. $ \mathbb{Z}/p \to (\mathbb{Z}/q)^*$, i.e. by an element $ i$ in $ (\mathbb{Z}/q)^*$ of order $ p$. thus $ G$ has the presentation $ \langle a , b : a^q \equal{} b^p \equal{} 1 , b a b^{\minus{}1} \equal{} a^i \rangle$. but it is easy to see that this group does not depend on $ i$. thus, up to isomorphism, there are exactly two grouos of order $ pq$.

[quote="-oo-"]but it is easy to see that this group does not depend on $ i$. thus, up to isomorphism, there are exactly two grouos of order $ pq$.[/quote]
And here you use, that $ (\mathbb{Z}/q\mathbb{Z})^\times$ is cyclic, so it has only one subgroup of order $ p$.
This follows from $ \mathbb{F}_q\equal{}\mathbb{Z}/q\mathbb{Z}$ being a field and every finite subgroup of the multiplicative group of a field is cyclic.


calculus
integration
logarithms
trigonometry
Euler
function
calculus computations

Use power series to compute $\int_0^{\frac{\pi}{2}}\ln \sin xdx$.

(It is easy to compute by adding a conjugate integral $\int\ln\cos xdx$, but I really have no idea about using power series...)

Use the relation:
\[
\Gamma(1+x/\pi)\Gamma(1-x/\pi)=\frac{x}{\sin x}
\]
to derive an expansion for $\ ln(\sin x)$. $ \Gamma(x)$ is the Euler function and the Taylor expansion of $\ln\Gamma(1+x)$ is given by
\[
\ln\Gamma(1+x)=-\gamma x+\displaystyle\sum_{k=2}^{+\infty}\frac{(-1)^k\zeta(k)}{k}x^k
\]
wiht $|x|< 1;\quad \gamma$ is the Euler number and $ \quad    \zeta(k)$ with  $\quad k\in\mathbb{N}$  the Riemann zeta function.


[quote="yunhua98"]Are there any new videos, cauase I watched a bunch and o more are coming out, I did more problems, still none.

P.S. Check out my website.[/quote]
You have to pass different categories of problems in order to get more videos.

And new videos (involving Geometry, Algebra) may be coming out with the new problems. (Just a guess)

Algebra is coming out late this year/early next year

I PMed rrusczyk quite a while ago, he said he's making more this fall.

[quote="XK-39"]I believe the following is the complete list of videos you can gain through Alcumus:
[url]http://www.artofproblemsolving.com/Alcumus/Videos/VideoList.php[/url]

As new videos come out, I think they are updated on that list. To when [i]exactly[/i] new videos come, I'm not quite sure.[/quote]
OMG, THANK U, ive been looking for these for a LONG TIME!!
thanks


5. Okay 22 liters = 29 grams. And 1 liter = 1000 cubic centimeters

So multiplying both sides of the second equation by 22, we get;

22 liters = 22,000 cubic centimeters
So the following must be true;
29 grams = 22 liters = 22,000 cubic centimeters
The amount of liters is completely useless at this point so we don't need it;
29 grams = 22,000 cubic centimeters
We can divide both sides by 22 and then write it as;

29/22 grams = 1000 cubic centimeters

A cubic meter is equivalent to 1,000,000 cubic centimeters.
So now we just need to find the amount of grams. Let the amount of grams in 1,000,000 cubic centimeters be $ g$.
Then that means that;

$ \frac{g}{1,000,000} \equal{} \frac{\frac{29}{22}}{1000}$
Multiplying by $ 1,000,000$ we get;

$ g \equal{} \frac{14,500}{11}$

So the amount of grams in one cubic meter is $ 14,500/11$.

I'll save number nine for somebody else who would like to solve it. Cuz I'm a nice person.  :P

[quote="gatorlover124"]5.In a room, each 22 liters of air weighs 29 grams. Each liter iss 1000 cubic centimeter. How much does a cubic meter of air in that room weigh? Express your answer to the nearest whole gram.

9.The sequence 2,3,5,6,7,10,11... contains all the positive integers from least to greatest that are neither squares nor cubes. What is the 400th term of the sequence?

Thanks to all of those who helped last time!!I really appreciate it :blush:[/quote]

9. Consider the first 425 numbers.

There are 20 squares, 7 cubes, and 1 of both.

Thus, 425 is the 399th number in our sequence.

Since 426 is not a square or cube, our answer is 426.

[quote="BOGTRO"]

There are 20 squares, 7 cubes, and 1 of both.

[quote]

theres more than one of both; 1,64...?

Yes. Refer to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=243445]this[/url].


induction
calculus
inequalities
function
limit
calculus computations

how do I use induction to show that a sub n is less than a sub n plus 1 for each n € natural numbers if a sub one is square root of 2 and a sub n plus 1 is the square root of 2 plus the square root of a sub n and they are all under the square root.

The second part is to show using induction that a sub n <2 for each n € natural numbers .

Also how would I show that for x in R: -1<x<1 and f(x) = x/(1-X^2) for x € I that f: I -R is one-to-one and onto and an example.

This is definitely not linear algebra. It's not calculus, either, but at least is has an inequality sign in there somewhere.

You really should learn a little $\text{\LaTeX}.$ Learn by copying templates and following examples. Whenever you see something and wonder how you say that, hold your mouse over the symbol or click on it; you'll see the codes that produced it. You won't see the dollar signs at the beginning and end; you'll just have to remember to always included those.

Here's what you said (I think):

Suppose that $a_1=\sqrt{2}$ and for $n\ge1,\,a_{n+1}=\sqrt{2+a_n}.$

Prove (by induction) that $a_n<a_{n+1}$  and $a_n<2$ for $n\in\mathbb{N}.$

Your second problem: if $I$ is the interval $(-1,1),$ then prove that the function $f$ given by the formula $f(x)=\frac{x}{1-x^2}$ is a one-to-one and onto function from $I$ to $\mathbb{R}.$

As for the first one, the inductive proof. Let's assume that we are trying to prove the compound statement with the "and" in it: $a_n<a_{n+1}$ and $a_n<2.$

First we have to verify this for $n=1.$ We have $a_2=\sqrt{2+\sqrt{2}}.$ Since $2+\sqrt{2}>2,$ this is clearly greater than $a_1=\sqrt{2}.$ On the other hand, we also clearly have that $a_1<2.$

Now let's work on the inductive step.

Assume (this is the inductive assumption) that $a_n<2$ and $a_n<a_{n+1}.$

Then $a_{n+1}=\sqrt{2+a_n}<\sqrt{2+2}=2.$

Also, $a_{n+1}=\sqrt{2+a_n}<\sqrt{2+a_{n+1}}=a_{n+2}.$

And we have proved our result by induction. Do you see where the inductive assumption was used in each line?

Now, for something that is calculus: We have just proved that this is a monotone bounded sequence. One of the most fundamental deep theorems of analysis is that every monotone bounded sequence in $\mathbb{R}$ converges to a real number. So we know that $\lim_{n\to\infty}=L$ exists.

Given that, what is $L?$ And do make direct use of the already-proved fact that the limit does exist.

(Please, no general statements about the behavior of function iterations - at least not until mikejones has a chance to look this over.)

Suppose you have an obviously continuous function defined on an interval, which we have. How to you prove that it is:

(1) one-to-one?

(2) onto?

(1) A continuous function on an interval is one-to-one if and only if it is strictly monotone - that is, either strictly increasing or strictly increasing. Thus one way to prove that such a function is one-to-one is to prove that it is monotone. You can do this directly with algebra and inequalities, or you can proved it by calculus, by computing the sign of the derivative.

(2) Proving that a function takes on every value in a certain range calls for the Intermediate Value Theorem.

In the case at hand, how does $f(x)$ behave as $x\to1^-$ or as $x\to-1^+$ ?


The arithmetic mean of three numbers is 86. What is the third number, given that the arithmetic mean of the first two is 81?

You can also think of it this way, the average number the first two numbers "pull" down from the arithmetic mean is $5$, therefore the total number the first two numbers "pull" down from the arithmetic mean is $5\times2=10$.  We must compensate for the $10$ unit "pull" (in other words, there is $10$ missing from the entire sum) therefore we must add $10$ to the mean which is $86+10=\boxed{96}$.

Problems like this can be done quickly if you think about it that way. :)


The sequence of integers is set as: $ a_1\equal{}2008^{2008}$; the value of $ a_{n\plus{}1}$-is sum of digits of number $ a_n$. Find $ a_5$


Dividing by 12, one typist can type 50 memos in 1 day. So three typists can type [b]150[/b] memos in one day.

[quote="GameBot"]Four typists can type a total of 600 memos in three days. At
that rate, how many memos can three typists type in one day?[/quote]

If four typists can type six hundred memos in three days, then they can type two hundred memos in one day, and each one can type fifty memos. This means three typists in one day can type $ 3 \times 50 \equal{} \boxed{150}$ memos.

EDIT: Beaten.


Let a_n=sum(k=2,n) n^(1/k) be given with n>=2. Compute lim(n-->oo) (a_(n+1)-a_n)

in English (qed)   :D
in Vietnamese (dpcm) = dieu phai chung minh  :D


calculus
analytic geometry
graphing lines
slope

I need help with a calculus problem.  Here it is.

A swimming pool is 12 meters long, 6 meters wide, 1 meter deep at the shallow end, and 3 meters deep at the deep end.  Water is being pumped into the pool at 1/4 cubic meters per minute, and there is one meter of water in the deep end.
wAt what rate is the water rising?

Assuming the pool has a uniform gradient, I don't think you need calculus at all. The dimensions of the surface of the water should be 6x6=36 (one half of the length since the sloped part of the pool is 1/2 full by height), so the water is rising by 1/144 metres per minute.

This thread belongs in Other Problem Solving Topics, or perhaps in some calculus-related topic in the advanced section.


Albatross bought a new dog and a new cat. Frankinfueter was very jealous, so he poisoned their food.

The first day the dog fainted 2 times. The second day the dog fainted 4 times. The third day he fainted 6 times.

The  first day the cat fainted 3 times. The second day the cat fainted 5 times. The third day it fainted 7 times.

At the end of 15 days how many more times did the cat faint than the dog?

This is very simple so please don't reply unless you are actually getting started in problem-solving.

[i]Note on moderator's edit: I changed the problem statement slightly for spelling and to make it less grotesque. Have a happy New Year.--tokenadult[/i]

Aww, tokenadult ruined it <_< .

[hide]

Each day, the cat faints one more time than the dog. After 15 days, the cat only fainted five more times than the dog because I poisoned the food with stronger poison at the end of the fifth day.

[/hide]


Soit x  un nombre réel et [x] représente sa  partie entière
Résoudre l’equoition  $[x]+[2x]+[4x]+[8x]+[16x]+[32x]=12345$.

Let $x$ be a real number, and let $[x]$ denote the greatest integer less than or equal to $x$.  Solve the equation $[x]+[2x]+[4x]+[8x]+[16x]+[32x]=12345$.

Put $x=n+q$ where $n$ is an integer part and $0\leqslant q<1$

Then

\begin{eqnarray*}[x]+[2x]+[4x]+[8x]+[16x]+[32x] &=& n+2n+[2q]+4n+[4q]+8n+[8q]+16n+[16q]+32n+[32q]\\ &=& 63n+[2q]+[4q]+[8q]+[16q]+[32q]\\&=& 12345\end{eqnarray*}

$q<1\implies 2q<2\implies [2q]\leqslant 1$

and similarly

$[4q]\leqslant 3, [8q]\leqslant 7, [16q]\leqslant 15, [32q]\leqslant 31$.

Summing that all up we get

$[2q]+[4q]+[8q]+[16q]+[32q]\leqslant 57$.

Therefore, $[2q]+[4q]+[8q]+[16q]+[32q]$ is the remainder when $12345$ is divided by $63$ and that remainder is not greater than $57$, but $12345\equiv 60\pmod{63}$, hence the equation has no solution.

I don't think there are any solutions.

Edit: Farenhajt beat me to it.

[hide="Proof"]On one hand, if $x \ge 196$, then \begin{eqnarray*}\lfloor x \rfloor+\lfloor 2x \rfloor+\lfloor 4x \rfloor+\lfloor 8x \rfloor+\lfloor 16 x \rfloor+\lfloor 32x \rfloor & \ge & 196+392+784+1568+3136+6272 \\ & = & 12348 \, . \end{eqnarray*} On the other hand, if $x < 196$, then \begin{eqnarray*}\lfloor x \rfloor+\lfloor 2x \rfloor+\lfloor 4x \rfloor+\lfloor 8x \rfloor+\lfloor 16 x \rfloor+\lfloor 32x \rfloor & \le & 195+391+783+1567+3135+6271 \\ & = & 12342 \, . \end{eqnarray*} [/hide]

yes, i did the way Ravi B did it but it's interesting the way Farenhajt used mods :)


Each point of the space is colored red or blue. Show that there exists a square with sidelength 1 such that either this square has (at least) 3 red vertices or this square has 4 blue vertices.

  Darij

I solved this one before. See :
http://www.mathlinks.ro/viewtopic.php?t=1486&highlight=red+blue

Pierre.

Thanks a lot! I am new to this topic...

  Darij

This problem is from the Bundeswettbewerb Mathematik (BWM) 1976. What else do you expect from a BWM seminar ?  :D BTW the BWM was established in 1970 in West Germany. I have the books for 1970-1997, later years can be found onto the net.  There are also some earlier years on the net but in non-official sources ( http://www.oliver-faulhaber.de/mathematik/bwm70.htm ). The other German national olympiad DEMO is better known in East Germany. For more information I can send you an English article on it. Don't hestitate to contact me.  :)

[quote="orl"]This problem is from the Bundeswettbewerb Mathematik (BWM) 1976.[/quote]

Thanks for the info, I didn't expect it, since on the seminar we were given a problem set where most of the problems were from BWM and they were marked with "BWM" and the corresponding year, but this one wasn't marked, hence I thought it was not from BWM. Yes, it's BWM 1976, Round 2, Problem 4.

  Darij


Let $ a,b,c>0$ . Prove that :

$ \frac {a(b \plus{} c)}{a^2 \plus{} 2bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} 2ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} 2ab} \geq 1 \plus{} \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}$

After using CS, we obtain an well-known result:
$ \sum\ \frac{1}{a^2\plus{}2bc} \geq\ \frac{2}{ab\plus{}bc\plus{}ca}\plus{}\frac{1}{a^2\plus{}b^2\plus{}c^2}$  :)

Can you demonstrate this result, nguoivn? Thanks in advance


$ \text{a,b,c are real,  } abc\equal{}1 \text{ and } a\plus{}b\plus{}c\le 0$

$ \text{prove that}$

$ \frac{1}{a} \plus{}\frac{1}{b}\plus{}\frac{1}{c} \minus{}2(a\plus{}b\plus{}c) \ge 9$

If $ a\equal{}b\equal{}\minus{}1$ and $ c\equal{}1$
$ abc\equal{}1$
$ a\plus{}b\plus{}c\equal{}\minus{}1 \le 0$
$ \frac{1}{a} \plus{}\frac{1}{b}\plus{}\frac{1}{c} \minus{}2(a\plus{}b\plus{}c)\equal{}1$
What is the original problem?


inequalities
IMO
algebra
maximization
IMO 1999
IMO Shortlist
Marcin Kuczma

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Really nice inequality.
The answer is $C=\frac{1}{8}$ and the equality holds when two $x_i$ are equal and

the rest are zero.
We have
$(\sum x_i)^4=(\sum x_i^2+2\sum x_ix_j)^2\geq 4(\sum x_i^2)(2\sum x_ix_j)= 8 \sum (x_ix_j\sum x_k^2)\geq 8\sum x_ix_j(x_i^2+x_j^2).$ The second ineq has

equality if and only if $n-2$ of $x_i$ are $0$. So WLOG assume that

$x_3=x_4=...=x_n=0$. Then for the first ineq to be equality we require that $x_1=x_2$
However that isclearly also sufficient for equality and we are done

Please stop copying entire solutions, Silouan, like you did with the problem about the hexagon, and also in this case ("your" solution comes straight from http://www.kalva.demon.co.uk/imo/isoln/isoln992.html, although you changed a few words).

Firstly I didnot say that this is ''my'' solution. And this is posted many days ago. And I consired good to post a solution. With this solution
i)someone could insrired to find another solution
ii) this problem doesnot seen as unsolved in the regional contests
iii) I remind this beautiful problem to all.

But if you want not me post such solutions then ok. It will not repeated. thanks for the hearing.

Silouanos

[quote="silouan"]There exist another solution for this.
We WLOG assume that $x_1+.....+x_n=1$ and we have $\sum_{i=1}^n {x_i^3(1-x_i)}\leq C$

Now let the function $f(x)=x^3(1-x)$ .

From Karamata inequality we have that $f(x_1)+....+f(x_n)\leq f(x_1)+f(1-x_1)+(n-2)f(0)\leq\frac{1}{8}$[/quote]

$f(x)=x^3(1-x)$
$f'(x)=3x^2 - 4x^3$
$f''(x)=6x-12x^2 \geq 0$ for $x \in \left [0, \frac{1}{2} \right ]$
it just concave for $x> \frac{1}{2}$

It suffices to prove it at the interval $[0,\frac{1}{2}]$.

the array $(x_1;x_2;x_3;...;x_n) \succ (x_1;1-x_1;0;...;0)$ , $f(x)$ is convex in $x \in \left [ 0, \frac{1}{2} \right ]$ then we have $f(x_1)+f(x_2)+...+f(x_n) \geq f(x_1)+f(1-x_1)+(n-2)f(0)$, did you reserve the inequality sign ?

[quote="orl"]Let $ n \geq 2$ be a fixed integer. Find the least constant $ C$ such the inequality
\[ \sum_{i < j} x_{i}x_{j} \left(x^{2}_{i} \plus{} x^{2}_{j} \right) \leq C \left(\sum_{i}x_{i} \right)^4
\]
holds for any $ x_{1}, \ldots ,x_{n} \geq 0$ (the sum on the left consists of $ \binom{n}{2}$ summands). For this constant $ C$, characterize the instances of equality.[/quote]
I will prove that $ C \equal{} \frac{1}{8}$. The inequality reduces to:

$ 8\left (\sum_{i\equal{}0}^{n}x_{i}\right )\left (\sum_{i\equal{}0}^{n}x_{i}^{3}\right )\le\left (\sum_{i\equal{}0}^{n}x_{i}\right )^{4}\plus{}8\sum_{i\equal{}0}^{n}x_{i}^{4}$

It's obvious when $ x_1\equal{}..\equal{}x_n\equal{}0$. Assume that not all are zero. Let $ y_i \equal{} \frac{x_i}{x_1\plus{}x_2\plus{}...\plus{}x_n}, i\equal{}1,2,..,n$. Then we have to prove that:

$ \sum_{i\equal{}1}^n 8y_i^3 \minus{} 8y_i^4 \le 1 \iff$

Consider $ f(x) \equal{} 8x^3 \minus{} 8x^4$. Then: $ f'(x) \equal{} 24x^2 \minus{} 32x^3$ $ f''(x) \equal{} 48x \minus{} 92x^2 \equal{} 48x(1\minus{}2x)$.

So $ f''(x) \ge 0 \forall x \in [0;\frac{1}{2}]$

Wlog assume $ y_1 \ge y_2 \ge y_3 \ge ... \ge y_n$.
Then: $ y_2,y_3,..,y_n \in [0;\frac{1}{2}]$. Since $ (y_2\plus{}y_3\plus{}..\plus{}y_n,0,0,...,0) \succ (y_2,y_3,...,y_n)$. We see that $ f(y_2\plus{}y_3\plus{}...\plus{}y_n) \plus{} f(0) \plus{} f(0) \plus{}.. \plus{} f(0) \ge f(y_2) \plus{} f(y_3) \plus{} .. \plus{} f(y_n)$. Let $ Z \equal{} y_2\plus{}..\plus{}y_n, Y \equal{} y_1$.

Then since $ f(0) \equal{} 0$ we just have to prove that:

$ f(Y) \plus{} f(Z) \le 1$ when $ Y\plus{}Z \equal{} 1$. But this is equivalent to:
$ 8(Y\plus{}Z)(Y^3\plus{}Z^3) \le (Y\plus{}Z)^4 \plus{} 8Y^4 \plus{} 8Z^4 \iff$
$ (Y\minus{}Z)^4 \ge 0$. So equality iff $ x_1\equal{}x_2, x_3\equal{}x_4\equal{}...\equal{}x_n\equal{}0$ or permutations. QED

[hide]I used just about three hours completing it, and I wrote about 2-3 pages LaTeX of notes/computations. [/hide]

[quote="huang"]Wait... if they say "fixed" $n$, doesn't that mean that $C$ can be in terms of $n$? If so, I got that $C=\frac{n-1}{n^{3}}$ (where the largest value that this can attain is precisely $\frac{1}{8}$[/quote]
Yes. $C$ should be in terms of $n$. If $x_1=x_2=1,x_3=x_4 =\cdots=x_n$ we get $2^4C \ge 2 \iff C \ge \frac{1}{8}$. So you have a mistake somewhere.

Can anyone look for a solution that uses this kind of approach?

For $n=2$ we have $(x_1+x_2)^4=({x_1}^2+{x_2}^2+2x_1x_2)^2 \ge 2\sqrt{({x_1}^2+{x_2}^2)2x_12x_2}^2=8x_1x_2({x_1}^2+{x_2}^2)$,so $C=\frac{1}{8}$ in this case.Now induct on $n$.

I did it with Lagrange multipliers.  The problem is equivalent to maximizing $f(x_1,...,x_n) = \displaystyle\sum_{i<j} x_i x_j(x_i^2 + x_j^2)  = \sum_i (1-x_i)x_i^3$, with the constraint that $\sum x_i = 1$.
Using the Lagrangian, at a maximum that isn't on the boundary $ 4x_i^3 - 3x_i^2 = 4x_j^3 - 3x_j^2\hspace {10pt} \forall i,j$.

$f(\dfrac 1 n, ..., \dfrac 1 n) = \dfrac {n-1} {n^3} < f(\dfrac 1 2,\dfrac 1 2, 0, ..., 0)$ for $n \geq 3$.

The function $g(x) = 4x^3 - 3x^2$ has a local maximum at 0 and a local minimum at $x=\dfrac 1 2$.

So the only other possibility for a maximum is a point of the type $(w,z,...,z)$ where $w > \dfrac 1 2, 1-w = (n-1)z$.

Let $n_1 = n-1$.

In that case, $f(w,z,...,z) = n_1(1-z)z^3 + n_1z(1-n_1z)^3$.

But $n_1(1-z)z^3 < (1-n_1z)(n_1z)^3$ for $n_1 \geq 2$, since $z < \dfrac 1 {2n_1}$.

So $f(w,z,...,z) < f(w,1-w)$.

But for $n=2$, at the maximum $4((1-w)^2+(1-w)w + w^2) = 3((1-w)+w)= 3$, which is only true if $w=\dfrac 1 2$.

So the max possible value of $f$ for any $n$ is $\dfrac 1 8$, at a point of the type $(\dfrac 1 2, \dfrac 1 2, 0, ..., 0)$.

[quote=orl]Let $n \geq 2$ be a fixed integer. Find the least constant $C$ such the inequality

\[\sum_{i<j} x_{i}x_{j} \left(x^{2}_{i}+x^{2}_{j} \right) \leq C
\left(\sum_{i}x_{i} \right)^4\]

holds for any $x_{1}, \ldots ,x_{n} \geq 0$ (the sum on the left consists of $\binom{n}{2}$ summands). For this constant $C$, characterize the instances of equality.[/quote]

[b]Answer.[/b] $C=\tfrac{1}{8}$.

(Proof) For equality, let $x_1=x_2=1, x_3=\dots=x_n=0$. Let us induct on $n \ge 2$. Now, for $n=2$ it is clear that $8ab(a^2+b^2) \le (a+b)^4$ for all $a, b \ge 0$. For $n>2$, pick $x_1, x_2$ to be the smallest two numbers in the set $\{x_1, \dots, x_n\}$. Fix $x_1+\dots+x_n=1$.

Then we have $x_1+x_2 \le \tfrac{2}{n} \le \tfrac{3}{4}$ for all $n>2$, so $$f(0, x_1+x_2, \dots, x_n)-f(x_1, x_2, \dots, x_n)=\left(3\sum_{k>2} x_k\right)x_1x_2(x_1+x_2)-x_1x_2(x_1^2+x_2^2)=x_1x_2(3-4(x_1+x_2)) \ge 0$$ hence we have reduced to the case of $n-1$ variables and induction applies. $\blacksquare$

Well at least I had fun while doing this,I started as
$$\left (\sum_{i=1}^n x_i\right)^4=\left(\sum_{i=1}^n x_i^2+2\sum_{i<j} x_ix_j\right)^2\leq 8 \left(\sum_{i=1}^n x_i^2\right)\left(\sum_{i>j} x_ix_j \right)$$
The rest is completely obvious,right?Well wrong ,i got distracted and went completely blank and discarded the whole thing altogether.

Now if we fix $\sum_{i=1}^n x_i$ we have to maximize the symmetric LHS and smoothing could really be magical here.Also note that :
$$x_1x_i(x_i^2+x_1^2)+x_2x_i(x_i^2+x_2)=x_i(x_1+x_2)(x_i^2+(x_1+x_2)^2-3x_1x_2)$$
The only summand that is left is $x_1x_2(x_1^2+x_2^2)$ but we can take care of him by making the $x_1,x_2$ two smallest numbers and then $x_ix_1x_2(x_1+x_2)$ would kill it easily,hence we just showed $F(x_1,x_2,...x_n)\leq F(x_1+x_2,x_3,...x_n)$ awesome right?The only case left is where there's nothing to take on the $x_1x_2(x^2+x_2^2)$ which is just the case $n=2$ and for that we easily have $C=\tfrac{1}{8}$.Note that we should clean out all the zeroes right away just to avoid some mess with equality cases.

So after reading the official solution i see one is smoothing and the other one is what i did at the beginning.

From setting $x_1=x_2$ in $n=2$, we get $C\ge 1/8$. Now we prove $C=1/8$ works. Now suppose $n>2$ and let $x_1\ge ...\ge x_n>0$ (note if $x_n=0$ then it reduces to the $n-1$ case). Let $S=\sum_{i<j} x_ix_j(x_i^2+x_j^2)$, and note that if we replace $(x_{n-1}, x_n)$ with $(x_{n-1}+x_n, 0)$, then $S$ increases by $(\sum_{i=1}^{n-2} 3x_{n-1}x_n(x_{n-1}+x_n)x_i))-x_{n-1}x_n(x_{n-1}^2+x_n^2)$, which is at least $0$ for $n>3$. Thus, this replacement makes $S$ increase, meaning we've reduced the problem to the $n=2$ case. But the $n=2$ case reduces to $(x_1-x_2)^4\ge 0$, so hence we are done.

The answer is $C =\frac{1}{8}$ and equality occurs when two variables are equal and the rest are $0$. Since $(a+b)^2 \ge 4ab$, we have
\begin{align*}
\left(\sum_{i=1}^n x_i \right)^4 &= \left( \sum_{i=1}^n x_i^2 + 2\sum_{i <j } x_ix_j \right)^2 \\
&\ge \left( 4 \sum_{i=1}^n x_i^2 \right) \left( 2 \sum_{i <j } x_ix_j \right) \\
&= 8 \sum_{k=1}^n x_k^2 \sum_{i<j} x_ix_j \\
&\ge 8 \sum_{i<j} (x_i^2+x_j^2) x_i x_j
\end{align*}
where the last inequality only occurs when $n-2$ of the variables are equal to $0$. In the case when $x_1 = x_2$ and $x_3 = \cdots = x_n = 0$, the inequality reduces to $(x_1-x_2)^4 \ge 0$.

The least such constant is $C=\frac{1}{8}$. By taking $n=2,x_1=x_2$, it is clear that $C\ge \frac{1}{8}$, hence it remains to show that $C=\frac{1}{8}$ always works.

By homogeneity, we may assume $\displaystyle\sum_{i=1}^n x_i=1$, hence we wish to show that
\[\sum_{1\le i<j\le n}x_ix_j(x_i^2+x_j^2) \le \frac{1}{8}\]
given that the sum of the $x_i$ is $1$. WLOG assume $x_1\ge x_2\ge \cdots \ge x_n$. We claim that if $n>2$, then we can increase that sum by replacing $x_1$ with $x_1+x_n$ and $x_n$ with $0$ as long as $x_1+x_n<3/4$. It is sufficient (and equivalent) to show
\[\sum_{1<i<n}\left(x_1x_i\left(x_1^2+x_i^2\right)+x_nx_i\left(x_n^2+x_i^2\right)\right)+x_1x_n\left(x_1^2+x_n^2\right) < \sum_{1<i<n}(x_1+x_n)x_i\left((x_1+x_n)^2+x_i^2\right)).\]
We explicitly compare the expressions in terms of $x_i$ to get that they have difference
\[(x_1+x_n)\left((x_1+x_n)^2+x_i^2\right))-\left(x_1\left(x_1^2+x_i^2\right)+x_n\left(x_n^2+x_i^2\right)\right)=\]
\[x_1^3+3x_1^2x_n+3x_1x_n^2+x_n^3+x_1x_i^2+x_nx_i^2-x_1^3-x_1x_i^2-x_n^3-x_nx_i^2 = 3x_1x_n(x_1+x_n)\]
times $x_i$. Then, it is sufficient to show
\[x_1x_n(x_1^2+x_n^2) < 3x_1x_n(x_1+x_n)\sum_{1<i<n}x_i.\]
Equivalently, it is sufficient to show
\[x_1^2+x_n^2 < 3(x_1+x_n)\sum_{1<i<n}x_i = 3(x_1+x_n)(1-x_1-x_n).\]
In fact, taking $x_1+x_n=a$ yields
\[3a(1-a)-x_1^2-x_n^2 \ge 3a-3a^2-a^2 = (3-4a)a > 0,\]
as desired. Then, we can continue iterating this "merging" operation until the sum of any two nonzero $x_i$ is at least $3/4$. Remark that if we had three nonzero $x_i$ at some point, the sum of the two minimal such $x_i$ would be at most $2/3$, hence there are only two nonzero $x_i$. Now, it suffices to show the inequality
\[x_1x_2(x_1^2+x_2^2) \le 1/8\]
for any nonnegative $x_1,x_2$ with sum $1$. Dehomogenize to see that it suffices to show the inequality
\[8a^3b+8ab^3 \le (a+b)^4.\]
However, this inequality is simply a rearrangement of the statement $(a-b)^4\ge 0$, which is trivially true. Hence, equality holds at
\[(x_1,\dots , x_n) = (1/2,1/2,0,\dots , 0)\]
and rearrangements thereof.

I claim that the answer is $C = \frac{1}{8}$ for all $n \geq 1$. Indeed, we can check that equality is achieved when $x_1 = x_2$ and $x_3 = \ldots = x_n = 0$ for all $n$.

There is nothing to prove for $n = 1$, and when $n = 2$, it is clear that $8ab(a^2 + b^2) \leq (a + b)^4$ for all nonnegative reals $a, b$ since\[8ab(a^2 + b^2) \leq 8ab(a^2 + b^2) + (a - b)^4 = (a + b)^4.\]When $n \geq 3$, we let $x_1 + x_2 + \ldots + x_n = 1$ because the inequality is homogeneous. Suppose we rearrange the numbers $x_1 \leq x_2 \leq \ldots \leq x_n$ into nondecreasing order since LHS and RHS are both symmetric.

We finish this problem with a bit of smoothing: note that if we replace $(x_1, x_2)$ with $(0, x_1 + x_2)$ then the LHS changes by $x_1x_2(3(x_1 + x_2)(x_3 + \ldots + x_n) - (x_1^2 + x_2^2)$ which can be obtained by a bit of expanding. If we let $x_1 + x_2 = t$, then this expression becomes\[x_1x_2(3t(1-t) - t^2 + 2x_1x_2) = x_1x_2(t(3 - 4t) + 2x_1x_2) \geq 0\]since $x_1 + x_2 \leq \frac{2}{n} \leq \frac{2}{3}$ for all $n \geq 3$.

Hence, this "smooth" does indeed increase (or rather, nondecresase) the LHS. We have reduced this to a $n-1$ variable inequality and then we can just induct down to $n = 2$ and get our desired result for all $n \geq 3$. $\blacksquare$

The answer is $\boxed{C = \frac{1}{8}}$ with equality holding at $(x_1,x_2,\dots,x_n) = (a,a,0,\dots, 0)$ for some non-negative $a$, and permutations.

[b][color=#f00]Claim: [/color][/b][Two-variable version] $x_1x_2(x_1^2+x_2^2) \leq \frac{1}{8} (x_1+x_2)^4.$
[i]Proof: [/i]Holds due to the expansion of $(x_1-x_2)^4 \geq 0$.

[b][color=#f00]Claim: [/color][/b][Smoothing] Suppose $n \geq 3$ and for each $i$ let $0 \leq x_i \leq x_{i+1}$. Replacing $(x_1,x_2)$ with $(0,x_1+x_2)$ never decreases the left-hand side of the inequality.
[i]Proof: [/i]Check this by some algebra.

Repeating smoothing the variables of our inequality, we may reduce the problem to the version where only $2$ of the variables are non-negative, which has already been proven. $\square$

We claim that the answer is $\boxed{C=\frac{1}{8}}$ for all $n$, and the equality case occurs when $x_1=x_2>0$ and $x_i=0, i>2$ or permutations, which clearly works. We prove that $C=\frac{1}{8}$ works, and that there are no other equality cases.

Because the desired inequality is homogenous, WLOG let $x_1+\ldots+x_n=1$. Notice that \begin{align*} \sum_{i<j} x_{i}x_{j} \left(x^{2}_{i}+x^{2}_{j} \right) &= \sum_{i<j}
\left(x_i^3x_j+x_j^3x_i \right) \\ &= \sum_{i<j} i^3j +\sum_{i>j} i^3j \\ &= \sum_{i=1}^n x_i^3(x_1+\ldots+x_{i-1}+x_{i+1}+\ldots+x_n) \\ &= \sum_{i=1}^n x_i^3(1-x_i). \end{align*} We want to prove that this expression is less than $C=\frac{1}{8}$, because we set $\sum_{i}x_{i}=1$.

[b][color=#f00]Claim 1:[/color][/b] Any optimal construction can have at most one number $x_i$ in the range $(0,0.5)$.
[i]Proof.[/i] Consider the function $f(x)=x^3(1-x)$. It is easy to see that $f''(x)=6x(1-2x)$, thus making the function convex on $(0,0.5)$. Now consider the following operation:

[b]Operation:[/b] Given two indices $i,j$ so that $x_i, x_j \in (0,0.5)$ and $x_i\le x_j$, decrease $x_i$ by a small amount $\epsilon$, and increase $x_j$ by the same amount, making sure that $x_i\ge 0, x_j\le 0.5$.

This configuration is higher than the previous configuration because $f(x)$ is convex on $(0,0.5)$. Therefore, an optimal construction cannot be operated on; thus it can only have at most value in $(0,0.5)$. $\blacksquare$

[b][color=#f00]Claim 2:[/color][/b] The optimal configuration contains at most two nonzero elements.
[i]Proof.[/i] Notice that $x_i=x_j=0.5$ is the only way to have more than one term at least $0.5$, and this contains two nonzero terms. This means that any other optimal construction could only have one value at least $0.5$. Because it can have at most one value in $(0,0.5)$, such a construction would have at most two nonzero values, proving the claim. $\blacksquare$

Therefore we are reduced to proving $x_ix_j(x_i^2+x_j^2)\le \frac{1}{8}$ given $x_i+x_j=1$. Notice that $x_i^2+x_j^2=(x_i+x_j)^2-2x_ix_j=1-2x_ix_j$, so letting $a=x_ix_j$ it suffices to prove that $$a(1-2a)\le \frac{1}{8}\Rightarrow -2\left(a-\frac{1}{4}\right)^2\le 0,$$ which is clearly true with equality only when $a=\frac{1}{4}$. Thus we have proven that $C=\frac{1}{8}$ works. Combined with $x_i+x_j=1$, this implies $x_i=x_j=\frac{1}{2}$. Taking off the condition that the sum of the $x_i$ is $1$, we get the complete equality case to be the one at the beginning of the proof.

The proof is complete.

ISL marabot solve

We claim the answer is $\boxed{\frac{1}{8}}$.

Part 1: Prove that the inequality always holds for $C=\frac{1}{8}$.
We have \[\left(\sum_i x_i\right)^4=\left(\sum_i x_i^2+2\sum_{i<j}x_i x_j\right)^2\ge 8\left(\sum_i x_i^2\right)\left(\sum_{i<j} x_ix_j\right)=8\sum_{i<j} \left(x_ix_j (x_1^2+x_2^2+\ldots+x_n^2)\right)\ge 8\sum_{i<j}x_ix_j(x_i^2+x_j^2),\]which proves the inequality always holds for $C=\frac{1}{8}$
$\blacksquare$.

Part 2: Characterize the instances of equality.
Note for equality to hold, we need \[\sum_{i<j}(x_ix_j(x_1^2+x_2^2+\ldots+x_n^2))=\sum_{i<j}(x_ix_j)(x_i^2+x_j^2).\]

Case 1: There exist $i$ and $j$ with $i<j$ so that $x_i, x_j>0$.
Note $x_ix_j(x_1^2+x_2^2+\ldots+x_n^2)=x_ix_j(x_i^2+x_j^2)$, so all other terms of the sequence are $0$. We have \[x_ix_j(x_i^2+x_j^2)=\frac{1}{8}(x_i+x_j)^4\implies 8x_i^3x_j+8x_j^3x_i=x_i^4+4x_i^3x_j+6x_i^2x_j^2+4x_ix_j^3+x_j^4\implies x_i^4-4x_i^3x_j+6x_i^2x_j^2-4x_ix_j^3+x_j^4=(x_i-x_j)^4=0\implies x_i=x_j\]

Case 2: There exists a unique $i$ so that $x_i>0$.
It's easy to see this doesn't satisfy equality in the original inequality (LHS is $0$ and RHS is greater than $0$).

Case 3: $x_i=0$ for all $i$.
This works.

So the instances of equality are $\boxed{x_i=x_j>0\text{ for some } i<j\text{ and everything else in the sequence is }0\text{ or }x_i=0\forall i}$
$\blacksquare$

Now we will show that no lower $C$ works. Checking the first equality instance gives \[2x_i^4\le16Cx_i^4\implies 2\le16C\implies \frac{1}{8}\le C\]
$\blacksquare$

We claim that the minimum value of $C$ is $\frac18$. Substituting $x_1=x_2=1$ and $x_3=x_4=\ldots=x_n=0$ gives $2\leq16C$, so $C\geq\frac18$. Now, we will prove that $C=\frac18$ works. Suppose that there are at least three nonzero $x_i$. Then, assume without loss of generality $x_1\geq x_2\geq x_3>0$. We claim that replacing $x_2$ with $x_2+x_3$ and $x_3$ with $0$ will increase the left hand side. We have
\begin{align*}
&x_2x_3(x_2^2+x_3^2)+(x_2^3+x_3^3)(x_1+x_4+\ldots+x_n)+(x_2+x_3)(x_1^3+x_4^3+\ldots+x_n^3)\\
&-(x_2+x_3)^3(x_1+x_4+\ldots+x_n)-(x_2+x_3)(x_1^3+x_4^3+\ldots+x_n^3)\\
=&x_2^3x_3+x_2x_3^3+(x_1+x_4+\ldots+x_n)(x_2^3+x_3^3-(x_2+x_3)^3)\\
=&x_2^3x_3+x_2x_3^3-3x_2x_3(x_2+x_3)(x_1+x_4+\ldots+x_n))\\
\leq&x_2^3x_3+x_2x_3^3-3x_1x_2x_3(x_2+x_3)\\
\leq&x_2x_3(x_2^2+x_3^2-(x_2+x_3)^2)\\
=&x_2x_3(-2x_2x_3)\\
<&0.
\end{align*}
Therefore, we only need to show $C=\frac18$ works when at most two of the $x_i$ are nonzero. This is equivalent to showing $x_1^4+4x_1^3x_2+6x_1^2x_2^2+4x_1x_2^3+x_2^4\geq8x_1^3x_2+8x_1x_2^3$, which is equivalent to $(x_1-x_2)^4\geq0$. Therefore, $C=\boxed{\frac18}$.

The answer is $C=\frac{1}{8}$. Equality holds when two of the values are equal and the rest are equal to $0$. The inequality is homogeneous, so WLOG set $\sum x_i=1$. Then we wish to find the least constant $C$ such that the inequality $\sum (x_i^3-x_i^4)\le C$ holds for all nonnegative $x_i$, or equivalently, maximize the value of $\sum (x_i^3-x_i^4)$. Now suppose that the answer is $C=\frac{1}{8}$ for all $2\le i\le n$. We show that the answer will be $\frac{1}{8}$ for $n+1$ (we will deal with the base case at the end of the proof). Clearly when $n$ or $n+1$ of the variables are zero, the whole sum becomes zero, which is not optimal. Otherwise, if we have anywhere from $1$ to $n-1$ zeros, the maximum value is $\frac{1}{8}$ by the inductive hypothesis. It suffices to show, then, that when none of the values are zero, we have $\sum (x_i^3-x_i^4)<\frac{1}{8}$.

It would be enough to show that there exist $x_i$ and $x_j$ that may be combined into $x_i+x_j$ such that the new sum (which has $n$ variables, and is therefore at most $\frac{1}{8}$) is greater than the old sum. Then we require
\[(x_i+x_j)^3-(x_i+x_j)^4>x_i^3-x_i^4+x_j^3+x_j^4\]
which simplifies to
\[3x_i+3x_j>4x_i^2+6x_ix_j+4x_j^2\]
so it would be enough that
\[3x_i+3x_j\ge 4(x_i+x_j)^2>4x_i^2+6x_ix_j+4x_j^2\implies x_i+x_j\le \frac{3}{4}.\]
Suppose that such $x_i$ and $x_j$ do not exist. Then for every pair of distinct $i,j$ we must have $x_i+x_j>\frac{3}{4}$ which implies that every $x_i$ is less than $\frac{1}{4}$, a contradiction. Therefore we may keep reducing the number of terms, which easily gives that the maximum occurs when $n-1$ of the variables are equal to zero while the other two are nonzero and sum to $1$.

Now, we wish to maximize $x^3-x^4+y^3-y^4$ subject to the condition $x+y=1$. This is equal to
\[-2x^4+4x^3-3x^2+x.\]
The derivative of this is equal to $(1-2x)^3$ which means that we need only to test $x=0,\frac{1}{2},1$. Clearly $x=\frac{1}{2}$ is optimal, which implies $y=\frac{1}{2}$ and gives $C=\frac{1}{8}$. Homogenizing, we find that equality occurs when two variables are the same and the rest are equal to $0$. We are done. $\blacksquare$

Let $x_1=x_2$ to get $2\le 16C$ so $C\ge \frac{1}{8}.$ We claim that $C=\frac{1}{8}$ is sufficient. We homogenize $x_1+x_2+\dots+x_n=1.$ The desired inequality becomes \[\sum_{i=1}^{n}{x_i^3(1-x_i)}\le \frac{1}{8}.\] Since $f(x)=x_i^3(1-x_i)$ is convex in $\left(0,\tfrac{1}{2}\right)$, we can apply Karamata's inequality: \[f(x_1)+f(1-x_1)+(n-2)f(0)\ge f(x_1)+f(x_2)\dots+f(x_n)\] which is the desired. This requires $(0.5,0.5,0,\dots)$, up to scaling or reordering.

Note that $(1,1,0,0, \dots , 0)$ gives that $C \ge \frac{1}{8}$. We prove that $C=\frac{1}{8}$ works. Note that $$\frac{1}{8}\left( \sum_{i}x_i \right) ^4 = \frac{1}{8} \left( \sum_{i}x_i^2+2 \sum_{i<j} x_ix_j \right)^2 \ge \sum_{i}x_i^2 \cdot \sum_{i<j} x_ix_j = \sum_{i<j} x_ix_j\left(\sum_{k}x_k^2\right) \ge \sum_{i<j} x_ix_j(x_i^2+x_j^2).$$ For equality to hold we need $x_ix_j\left(\sum_{k}x_k^2\right)=x_ix_j(x_i^2+x_j^2)$. This is obviously false if there is more than $2$ nonzero $x_i$'s by taking $x_i$ and $x_j$ to be nonzero. If there is at most one nonzero $x_i$ we have that $(0,0, \dots 0)$ works. If there is $2$ nonzero $x_i$'s then we get the solution $(a,a,0,0, \dots , 0)$ and permutations.

We claim that the answer is $\frac{1}{8}$. By taking $x_1=x_2=1/2$ and everything else 0, we can see that no smaller $C$ works. We will then show that 1/8 works.\\

Since the inequality is homogenous, we can WLOG that $x_1+x_2\cdots + x_n=1$. Then, the inequality becomes $$\sum_{i=1}^n x_i^3(1-x_i)\leq \frac{1}{8}.$$

Note that $$f(x)=x^3-x^4$$ is convex in the interval $[0,1/2].$\\

We claim that if $$\sum_{i=1}^n x_i^3(1-x_i)$$ is maximized, then we cannot have 3 or more nonzero variables. Suppose FTSOC that the sum is maximized while having three nonzero variables. Then, we can find two variables that are both strictly between 0 and 1/2, since there are three nonzero variables. However, since $f$ is convex on $[0,1/2]$, we can "push" those two chosen variables away from each other until one of them reaches 0 or 1/2. Due to Karamata's, this pushing increases the value of the sum, contradiction.\\

Therefore, we can reduce this to 2 variables instead of 4, in which case it is easy to see that both being $1/2$ is optimal, giving the equality case. Undoing the homogenization, equality occurs when two variables are equal and the rest are all 0.

I tried solving the problem by simplifying the LHS. This gave the following problem: Find the minimum value of C such that f(x1)+f(x2)+.....f(xN) <= c for all integers N>1 and positive reals x1, x2, .......xN summing to 1, where f(x) = x^3-x^4. I figured out that f is convex from x=0 to x=1/2, and used Jensen's inequality to figure out that, if x1,x2,.....xN are all at most 1/2, then the LHS is at least Nf(1/N), and if N=1/2, then Nf(1/N)  = 1/8. I'm trying to find a maximum value but this only leads to a set of minimum values. Did i use the wrong method or is there a way to solve this with Jensen's?

We claim $C=1/8$. Let $x_1=x_2 >0$ and $x_3=x_4=\dots=x_n=0$ to get $C\geq \frac 18$. The inequality is homogenous, so let the sum of the variables be $1$. It remains to show $\sum x_i^3-x_i^4 \leq \frac 18$, but this follows by Karamata, as $x^3-x^4$ is convex over the interval $\left( 0, \frac 12\right)$ and $\left(x_1, 1-x_1, 0, \dots, 0 \right)$ majorizes $(x_1, x_2, \dots, x_n)$. Equality is when $x_1=x_2 >0$ and the rest of the variables are $0$.

There is an SOS expression.
\[\left(\sum_{i=1}^na_i\right)^4-8\sum_{1\le i<j\le n}a_ia_j\left(a_i^2+a_j^2\right)=\left(\sum_{i=1}^na_i^2-2\sum_{1\le i<j\le n}a_ia_j\right)^2+8\sum_{\substack{1\le k\le n,\\1\le i<j\le n,\\k\ne i,j}}a_k^2a_ia_j.\]

Answer is $C = \frac 18$.

The idea is to do smoothing (or Karamata if you like high-tech stuff.) Fix $\sum_{i=1}^n x_i = 1$, and consider the change $(x_1, x_2) \to (0, x_1+x_2)$. This alters the LHS by
\begin{align*}
A &= x_1 \sum_{j=2}^n x_j(x_1^2+x_j^2) + x_2\sum_{j=3}^n (x_2^2+x_j^2) - (x_1+x_2) \sum_{j=3}^n x_j((x_1+x_2)^2+x_j^2) \\
&= 3(1-x_1-x_2)(x_1+x_2) - x_1^2-x_2^2.
\end{align*}
One can check that this is strictly greater than zero for $x_1 + x_2 < 1$.

Thus we can reduce the problem to the case where $x_1 = x_2 = k$ and all the rest of the $x_i$ are $0$. This yields $C = \frac 18$.

Wow, this was very technically involved.

We claim the minimum is achieved at $C = \frac{1}{8}$, with equality at $x_1 = x_2 = \frac{1}{2}$ and $x_3 = x_4 = \dots = x_n = 0$.

Homogenize such that $\sum x_i = 1$. Then we wish to find the least constant $C$ such that,
\begin{align*}
    \sum_{1 \le i < j \le n} x_i x_j (x_i^2 + x_j^2) \leq C
\end{align*}
However the left hand side is simply,
\begin{align*}
    \sum_{1 \le i < j \le n} x_i x_j (x_i^2 + x_j^2) &= \sum_{i} x_i^3(1 - x_i)\\
    &= \sum_i x_i^3 - \sum_i x_i^4.
\end{align*}
Thus we wish to find the least constant $C$ such that,
\begin{align*}
    \sum_i (x_i^3 -x_i^4) \leq C
\end{align*}
It suffices to maximize the left hand side. Note that the function $f(x) = x^3 - x^4$ is convex on the interval $\left(0, \frac{1}{2}\right)$ and that there is at most one $x_i > \frac{1}{2}$, which we assume wlog is $x_1$. Now Karmata's gives,
\begin{align*}
    f(x_1) + f(1- x_1) + (n-2) \cdot f(0) &\geq \sum f(x_i)\\
    f(x_1) + f(1- x_1) &\geq \sum f(x_i)
\end{align*}
Thus we wish to maximize,
\begin{align*}
    x^3 - x^4 + (1-x)^3 + (1-x)^4.
\end{align*}
However the derivative of this is simply $(1-2x)^3$ from which we finish, with equality achieved at $x = \frac{1}{2}$.

I claim that the answer is $C = \frac{1}{8}$. Note that $n = 2$ and $x_1 = x_2 = 1$ implies $C \geq \frac{1}{8}$.
To show $C = \frac{1}{8}$ works, just note
\[ \left( \sum x_i\right)^4 = \left( \sum x_i^2 + 2\sum x_ix_j \right)^2
	\stackrel{AM-GM}{\ge} \left( 2\sqrt{2} \left( \sum x_i^2\right) \left( \sum x_ix_j\right) \right)^2
= 8 \left( \sum x_i^2\right) \left( \sum x_ix_j \right) \]
One can easily check that the last expression is at least $8 \sum (x_i^2 + x_j^2)x_ix_j$. Note that equality holds only if
\[ \sum x_i^2x_jx_k = 0 \]
which implies that all $x_i$ are zero except for say $x_1$ and $x_2$. But then $x_1^2 + x_2^2 = 2x_1x_2$ implies that $x_1 = x_2$. Since $x_1 = x_2$ and $x_3 = \dots = x_n = 0$ also implies equality, this is the only equality case. $\blacksquare$

$\left(\sum x_i \right)^4 = \left(\sum x_i^2 + 2\sum x_ix_j \right)^2 \geq  8\left( \sum x_i^2 \right)\left(\sum x_ix_j \right)$ $= 8\left(x_jx_k \left(\sum x_i^2\right)\right) \geq \sum x_{i}x_{j} \left(x^{2}_{i}+x^{2}_{j} \right)$, equality holds iff $n-2$ of them are $0$ and other $2$ are equal.

We claim that $\min(C)=\frac{1}{8}$. Plugging in $x_1=x_2$ for $n=2$ yields that $C \geq \frac{1}{8}$ so we cannot do better. Now, we try to prove the following inequality,
\[\sum_{1\leq i < j \leq n}x_ix_j(x_i^2+x_j^2)\leq \frac{1}{8} \left(\sum_{1\leq i \leq n}x_i\right)^4\]
We immediately observe that the given inequality is homogeneous and thus, assume WLOG $x_1+\dots +x_n=1$. Then, we smooth this inequality. Assume $0<x_1\leq x_2 \leq \dots \leq x_n$. Consider smoothing as follows. We simply do the transformation $(x_1,x_2) \mapsto (0,x_1+x_2)$. This changes the LHS by,
\[\sum_{i=3}^n x_i(x_1+x_2)((x_1+x_2)^2 + x_i^2) - \sum_{i=3}^n x_1x_i(x_1^2+x_i^2) - \sum_{i=3}^n x_2x_i(x_2^2+x_i^2) - x_1x_2(x_1^2+x_2^2)\]
Simplifying this expressions yields,
\begin{align*}
S &= (x_1+x_2)^3(1-x_1-x_2)-(x_1^3+x_2^3)(1-x_1-x_2)-x_1x_2(x_1^2+x_2^2)\\
&= x_1x_2\left(3(x_1+x_2)(1-x_1-x_2)-x_1^2-x_2^2\right)
\end{align*}
Now, we wish to show that this expression is positive for $x_1+x_2\leq 1$. Simply note that,
\[S= x_1x_2((x_1+x_2)(3-4(x_1+x_2) +2x_1x_2)\]
Thus, it is easy to see that $S>0$ when $x_1+x_2\geq \frac{3}{4}$. But since $x_1,x_2$ are the minimal terms, we know that $x_1+x_2 \leq \frac{2}{3}<\frac{3}{4}$. So $S$ is indeed positive as desired.\\
Now, we can ignore the newly created terms of value zero and are reduced to the $n-1$ case. Thus, we are reduced to the $n=2$ case via induction.\\

Here, we smooth again. Let $\epsilon=\frac{x_2-x_1}{2}$. We simply do the transformation $(x_1,x_2)\mapsto (x_1+ \epsilon, x_2-\epsilon)$. Then, it is easy to see that
\[(x_1+\epsilon)(x_2-\epsilon)((x_1+\epsilon)^2+(x_2+\epsilon)^2) > x_1x_2(x_1^2+x_2^2)\]
when $\epsilon>0$. Thus, we are left with $x_1=x_2$ at which point the given inequality is obvious.\\

Thus, the desired inequality holds for all non-negative reals $x_1,x_2,\dots,x_n$ for any natural number $n\geq 2$, with equality holding if and only if $x_1=x_2=k$ for some $k\in \mathbb{R}_{\geq 0}$ and $x_3=x_4=\dots =x_n=0$.


What is the bond angle of $H-C-OH$ in phenylmethanol ?

i dont't know exactly how to find the angle...i can guess you shoul consider the effects,or RMN,or the hybridisation...
i know that there is a freeware software that can calculate the angles between the bonds for you...
it is ChemSketch.I dont have time right now but you can try and see f it works..
Good luck!!

it should be around 109.5
the benzillic carbon is sp3, although the presence of sp2 carbon of benezene group might affect it, i doubt it would be much.


probability
calculus
calculus computations

Can you help me with this question please?

Derive the relationship between nominal and effective interest rate if constant nominal interest rate r is paid continuously.

thanks

This isn't probability by the way

If Y is the amount of money at time t, then

$ Y' \equal{} Yr$
so
$ Y \equal{} e^{rt}$

This means that in any given unit of time, the interest rate over that unit of time was
$ e^r\minus{}1 \equal{} r \plus{} \frac{r^2}{2} \plus{} ...$

If you're using percentages, subtract one and multiply by a hundred.


analytic geometry
geometry
angle bisector
similar triangles

A circle of radius 1 rolls around the inside of a 6-8-10 triangle. How far does the center of the circle travel?

Suppose C is the right angle, B is the larger angle, and A the smaller angle. Bisect A,B and let them intersect the opp. side at D,E. Now we want to measure the length between the centre of the circle when it is at B and at C. By the angle bisector theorem, EC=3. Therefore by similar triangles, the point of tangency on BC of the cricle is 2 from B, and is obviously 1 at C so the total distance travelled up that side is 3. We can do the same for between C and A, and we find that it travels 4. Clearly, since the locus is a right-angled triangle, between AB it travels 5, so the total distance = 12.

[hide]I had to use Cartesian coordinates, and equation for distance from point to line, the center follows a 3-4-5 triangle, making the distance 12.[/hide]


HEY ANAKIN, COME BACK TO THE LIGHT SIDE OF THE FORCE.

The light side of the force.... that is something rather subjective.... :P
Now, really:

Neniu parolas esperanton?

[quote="Dave Vather"]Hello!! Do you speak Esperanto?
If you do then...
Kiel vi estas?[/quote]
 :rotfl:
Lol, Esperanto.
I hope to maybe learn that someday.

[quote="me@home"]Wasn't esperanto created as the ultimate language or something?
And from a little linguistic knowledge, you just said "anybody speak esperanto"
"no speako eseperanto"[/quote]
I think "who are you" and "no one speaks esperanto?"

I'm coming to this conversation late (and, in any case, was just looking around to see if anything interesting was said about the AMC contests...) -
this has nothing to do with math  - but I can't resist adding a few comments, since I know a few Esperanto speakers.

(A) Some people do indeed speak Esperanto! (Kvankam mi malofte havas okazon por paroli ĝin,  foje mi uzas la lingvon interrete.) I've even translated a few simple math problems into Esperanto. It has a slight aura of crackpottery about it, but math people usually can overlook that.

(B) it wasn't intended to be the "ultimate" language, but an easy-to-learn language that everyone could use as a common second language -- with the added benefit of being, in some sense, "neutral", not giving native speakers any advantage.  This is pretty unlikely to ever happen on a large scale, of course -- but it is at least true that the language *is* easy to learn and can be used to express pretty much anything you want to say. Books are published in it; conferences are held using it; people have even met, fallen in love,  and married, sharing only Esperanto as their common language.  Currently,  of all the languages used at wikipedia.org, Esperanto is 15th, with over 65000 entries -- this is behind Dutch, Finnish, and Norwegian, but ahead of Slovak, Czech, Danish, and Romanian, for example.

There's a free ten-lesson Esperanto course for English speakers you can find online -- it's remarkable that in just a week or so, you can gain enough passive understanding of a language to read it fairly well.

(oh, and "kiel vi estas?" is HOW are you, not who are you. You got the other one right.)

The free lesson is amazing.
[url]http://pacujo.net/esperanto/course[/url]
I have started right away.

The thing with linguistic domination is that there's always a snowball-effect : the more people who learn it/speak it, the more other people would be interested to learn it as well, at the cost of other languages.
English is dominant now and all the anglophones have to do is firmly refuse to learn Esperanto or whatever and their position is not treathened at all.

Apart from that, I might give Esperanto a chance some day. :wink:   A language without a lot of exceptions (no genders, no difficult conjugations,...) and with a decent spelling is so nice.

Ah I am open to new languages.
Though bengali and english will rock forever, the others should get some chance as well.

Esperanto when it was founded replaced Latin as the universal language.  They both had the same problem: no one spoke them.


What is the coefficient of $ a^{2}b$ in the expansion of $ (a\plus{}b)^{3}$?

[hide="Solution"]

The expansion of $ (a\plus{}b)^3$ is:

$ a^3\plus{}3a^2b\plus{}3ab^2\plus{}b^3$

Therefore, the coefficient of $ a^2b$ is $ \boxed{3}$.

[/hide]

Or, by the Binomial Theorem, the coefficient is $ \binom{3}{1}(1)(1)\equal{}\boxed{3}$.


quadratics
calculus
integration
Ring Theory
advanced fields
advanced fields theorems

Let n=pm , where p is an odd prime number and m >= 1. Let j  be a primitive n-th root of unity, L=[b]Q[/b](j), K=[b]Q[/b](j+j -1 ), so that L is the n-th cyclotomic field and K is its maximal real  subfield. It is well-known that the prime number p is totally ramified in L, that is p=P:phi:(n), where P is a prime ideal of  L. From this we can see that p is also totally ramified in K, hence p=Q:phi:(n)/2, where Q=P:cap:K is a prime ideal of K. Obviously, Q=P2.

:roll: Could anyone please prove that the relative discriminant of the quadratic extension L|K is the prime ideal Q?

This claim is made in the article "On the integral basis of the maximal real subfield of a cyclotomic field", by Joseph J. Liang, published on J. Reine Angew. Math. (1974).

I've already solved it!

Since P is the only prime of L that is ramified in L|K, the relative different D of this extension is a power of P, say Ps. Now, let B denote the ring of algebraic integers of L. Clearly, char(B/P)=p, which does not divide the ramification index of P in L|K (this is 2). Thus s=2-1=1. So we proved that D=P.
The relative discriminant d of the extension L|K is N(D), where N denotes the relative norm of L|K. Finally, we note that d2=N(D)2=N(D2)=N(P2)=N(BQ)=Q2.
Hence d=Q.


induction
number theory unsolved
number theory

Let be $ n\in \mathbb{N}$ with $ n \ge 2$ . Prove that $ \bigvee n$ , the equation $ x_1^2 \plus{} x_2^2 \plus{}...\plus{}x_n^2\equal{}nx_1x_2...x_n$ has an infinity of solutions $ (x_1,x_2,..,x_n)$ with $ x_1,x_2,...,x_n \in \mathbb{N^*}$ .

It is the makov problems.
You can solve as follow with  strategies.
Easy to check $ x_1\equal{}x_2\equal{}...\equal{}1$ is a solution of this equation.
Let $ x_{1}\equal{}\min\{x_1,...,x_n\}$
$ x_1^2\minus{}nx_1x_2...x_n\plus{}x_1^2\plus{}...\plus{}x_n^2\equal{}0$
because it has a root $ x_1$ so this equation has other solution :
$ x_1'\equal{}nx_2...x_n\minus{}x_1>x_1$
Apply this congress for infinite time then the equation has infinite solution.

Thank you very much,[b] TTsphn[/b], for your help , but sincerily I didn't understand your ideea , here :
[quote="TTsphn"]Apply this congress for infinite time then the equation has infinite solution.[/quote] . I mean , how do you apply this for infinite times ? Could you be a little bit more explicit,please ?...

[color=blue]I just wonder if there isn't another proof.Maybe one using [b]mathematical induction[/b].If somebody has one,please post it here :) [/color]

If $ (x_1,x_2,x_3,...,x_n)$ is solution then $ (nx_1...x_n\minus{}x_1,x_2,...,x)n)$ is too.
And use this progress for each root you obtain it can represent an other root and it is increasing.


Find the area of the figure $\left|x^{2}-|y|\right|+\left|y^{2}-|x|\right|+\left|x+y\right|=a^{2},a\in\mathbb{R}$

Here's a plot for $a=0,\frac{1}{2},1,2,3,4$ from http://www.quickmath.com:

[url=http://ezimagecenter.com/out.php/i155_AreaofaFigure.JPG][img]http://ezimagecenter.com/out.php/t155_AreaofaFigure.JPG[/img][/url]