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This problem came up at school as a consequence of a problem we were assigned.

Prove that every natural number can be expressed as a sum or difference of powers of 3. (eg, 25 is 27-3+1, 40 is 27+9+3+1, 5 is 9-3-1, 47 is 81-27-9+3-1)

I have a feeling it has something to do with number theory, but that's never been my strong point.   :maybe:

In ternary, there are three possibilities for each "digit".

Sum and difference works the same way, just shifted one down.

For each power of 3, you have three options: add it, don't use it, or subtract it. Clearly, this gets the job done because every number can be expressed in ternary.

I'm going to work on the rigorous proof, if that's what you want.

This reminds me a lot of the [url=http://en.wikipedia.org/wiki/Balanced_ternary]balanced ternary[/url] system.

its very very easy
number $ n$ can be written in this form:
$ n\equal{}\underbrace{3^0\plus{}3^0\plus{}\ldots \plus{}3^0}_{n\ times}$.
But if there is a restriction... i have to see xD

[quote="pelao_malo"]its very very easy
number $ n$ can be written in this form:
$ n \equal{} \underbrace{3^0 \plus{} 3^0 \plus{} \ldots \plus{} 3^0}_{n\ times}$.
But if there is a restriction... i have to see xD[/quote]
I think the original poster means that each power of 3 must be distinct.

I have an "answer."
Each power of three can be written with a + or - sign when chosen.  There are thus 2^n choices of defining these functions.  Furthermore, 2= -1 mod 3, and since 1 is included as a power of three, this [b]somehow [/b]proves that each integer n can be written as a sum or difference of a power of three.

Or it could not be used at all. That's the third option. :)

[quote="Yongyi781"]In ternary, there are three possibilities for each "digit".

Sum and difference works the same way, just shifted one down.

For each power of 3, you have three options: add it, don't use it, or subtract it. Clearly, this gets the job done because every number can be expressed in ternary.[/quote]

I'm not sure the ternary idea necessarily works.  Every integer can be expressed in base 4, base 5 etc. as well as ternary.  This does not necessarily imply that those powers can be used to express all numbers.  1 and 4 can't make 2.  1 and 5 can't make 2.

The reason why it works in base-3 and not in base-4 is that in balanced ternary, we only need digits: -1, 0, and 1. In, say, base-4, for example, we'd need four digits, but if we are limited to -1, 0, and 1, we can't do it.

No, but there is a one-to-one correspondence between balanced ternary and ternary. [url=http://en.wikipedia.org/wiki/Balanced_ternary]This article[/url] mentions how to convert between balanced ternary and ternary.

You can also use [b]induction[/b]. Here's an outline of the major steps in one possible approach:

► Let the number be n, and let m be the nearest power of 3
► Show then that |m-n|<= m/3
[i]This shows that if we set aside m and want to represent the difference n-m, we will only need to use lower powers of 3[/i]
► Observe that there are a finite number of integer powers of 3 <= m
[i]This shows that the algorithm must terminate[/i]

The simplest method is as follows: rewrite the number in ternary.  For example, $ 34$ becomes $ 1021$.  Note that any digit of $ 2$ can be written as $ 2\times 3^n$.  This can also be written as $ 3\times 3^n \minus{} 1\times 3^n$ or $ 3^{n\plus{}1} \minus{} 3^{n}$.  By this method, the $ 2$ becomes a $ \minus{}1$ and the previous digit is shifted up by one; if it was a $ 0$ it becomes a $ 1$, if it was a $ 1$ it becomes a $ 2$ and we can repeat this process, if it was a $ 2$ it becomes a $ 0$ and the previous digit is shifted up one.  In any case, this method may be repeated until we have reached a balanced ternary number.  Then rewrite it as the sum and difference of the powers of $ 3$ in the number.

HEHEHE, generating functionology!

$ (1/x\plus{}1\plus{}x)\equal{}1/x\plus{}1\plus{}x$
$ (1/x\plus{}1\plus{}x)(1/x^3\plus{}1\plus{}x^3)\equal{}1/x^4\plus{}1/x^3\plus{}1/x^2\plus{}1/x\plus{}1\plus{}x\plus{}x^2\plus{}x^3\plus{}x^4$
Now, simple induction shows us that

$ (1/x\plus{}1\plus{}x)(1/x^3\plus{}1\plus{}x^3)...(1/x^{3n}\plus{}1\plus{}x^{3n})$ Gives each term between 1/x^(1+3+9+...+3n) to x^(1+3+9+...+3n) exactly once.

Now, The power of x represents the natural number to be represented, and each term on the L.H.S. represents either adding or subtracting a power of 3.

Now, obviously every number can be represented as just shown, and it is unique, since no matter how many powers of 3 we are allowed to use (corresponding to L.H.S. terms) we have only one term on the R.H.S. for it.

Coolest way ever.

rofler's technique is also a concise way to prove that binary representations and so forth are unique:  note that

$ (1 \plus{} x)(1 \plus{} x^2)(1 \plus{} x^4)...(1 \plus{} x^{2^n})) \equal{} 1 \plus{} x \plus{} x^2 \plus{} ... \plus{} x^{2^{n \plus{} 1} \minus{} 1}$

which can be proven by multiplying both sides by $ 1 \minus{} x$.  Similarly,

$ (1 \plus{} x \plus{} x^2)(1 \plus{} x^3 \plus{} x^{6})(1 \plus{} x^9 \plus{} x^{18})...(1 \plus{} x^{3^n} \plus{} x^{2 \cdot 3^n}) \equal{} 1 \plus{} x \plus{} x^2 \plus{} ... \plus{} x^{3^{n \plus{} 1} \minus{} 1}$

which can be proven by multiplying both sides by $ 1 \minus{} x$.  rofler's identity follows by dividing the above by $ x^{\frac { 3^{n \plus{} 1} \minus{} 1}{2} }$; unique balanced ternary representation is therefore a direct consequence of unique ternary representation.


I remember that $k=\sqrt{\sqrt6}$,then it would be easier.


is there a way to use induction to prove the sum of squares formula? $ 1+4+9....n^{2}=\frac{n(n+1)(2n+1)}{6}$

[hide="HINT"]
See if it is true for $ n=1$
Assume true for $ n=k$
Prove it is true for $ k+1$.
[/hide]

Why do you post problems that have been solved already?

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=161354#903587[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42048#265822[/url]

And if someone wants to do it without induction:
[hide="hide"]
Look in my 2nd post in this thread:
http://www.mathlinks.ro/Forum/viewtopic.php?p=900576#900576
[/hide]

That does work; I never thought of doing it that way!

So there's actually a really neat way of doing this that can be demonstrated with pictures (I love pictures).
It involves the fact that the sum of the first $ n$ odd numbers equals the $ n$th square.

The concept is summarized below:

[hide][i]...Image may take a few moments to load...[/i]
[img]http://users.wpi.edu/~anirvan/Linked/demo.gif[/img]

And this gives you:
$ \sum_{k=1}^{n}k^{2}= \sum_{k=1}^{n}(2k-1)(n-k+1)$
$ \iff \sum_{k=1}^{n}k^{2}= \frac{n(n+1)(2n+1)}{2}-2 \sum_{k=1}^{n}k^{2}$
$ \iff \sum_{k=1}^{n}k^{2}= \frac{n(n+1)(2n+1)}{6}$[/hide]


Let $ a,b,c$ be real numbers such that $ a^2\plus{}b^2\plus{}c^2\equal{}9$. Show that the inequality $ 2(a\plus{}b\plus{}c)\minus{}abc \le 10$ holds.

[quote="micf"]Let $ a,b,c$ be real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 9$. Show that the inequality $ 2(a \plus{} b \plus{} c) \minus{} abc \le 10$ holds.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=52
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122166


[quote="ckck"]This was in a mathcounts test.[/quote]
Are you sure you typed it up right and that it wasn't the three that was squared? Because this problem isn't mathcounts level...

You typed it up wrong, Preet was right. The actual problem is $(3\cdot7^{n})^{2}=63$

Divide by 9, then you get $7^{2n}=7\rightarrow n=1/2$

[quote="13375P34K43V312"][hide="are you sure this is the right level?"]
$7^{2n}=21$
$2n=\log_{7}21$
$n=2\log_{7}21$
$n=\log_{7}441$
$n=\frac{\ln{441}}{\ln{7}}$

Then type it into your calculator. [/hide]
no, it's not.[/quote]

[hide]In the third step, since you're dividing each side by two, shouldn't it be

$n=\frac{\log_{7}21}{2}$?

That would change the answer.[/hide]

Oh, yeah, sorry. Well it doesn't matter anyway
Note that your expression is equivalent to $\log_{7}\sqrt{21}$.


inequalities
algebra
polynomial
inequalities proposed

(stupid post deleted)

I'm sorry, I don't know what I did, probably switch the sign or something.

[quote]
I don't think it works, try 0.1,0.1,10

It is true though if  are the sides of a triangle.
[/quote]

Hi, SplashD,

Even though I have two different solutions to this problem, I did try as you suggested and it [b]works[/b] perfectly (I also tried the easier tuple $(1,1,100)$).

I like too much this inequality  :)  :wink:

I tried to solve it using

[quote="toanhocmuonmau"][quote="ductrung"]For any positive real numbers $ a,b$ and $ c$,
\[ \frac{1}{a^{2}+bc}+\frac{1}{b^{2}+ca}+\frac{1}{c^{2}+ab}\ge\frac{3(a+b+c)^{2}}{2(a^{2}+b^{2}+c^{2})(ab+bc+ca)}\]
[/quote]
[hide="hint"]
The inequality is equivalent to
\[ \sum\frac{b^{2}+c^{2}-bc}{a^{2}+bc}+3 \ge \frac{3(a+b+c)^{2}}{2(ab+bc+ca)}\]
By Cauchy Schwarz inequality, we have
\[ \sum\frac{b^{2}+c^{2}-bc}{a^{2}+bc}\ge \frac{(2(a^{2}+b^{2}+c^{2})-ab-bc-ca)^{2}}{\sum (a^{2}+bc)(b^{2}+c^{2}-ca)}=\frac{(2(a^{2}+b^{2}+c^{2})-ab-bc-ca)^{2}}{(ab+bc+ca)(a^{2}+b^{2}+c^{2})-4abc(a+b+c)}\]
Hence, it suffices to prove that
\[ \frac{(2(a^{2}+b^{2}+c^{2})-ab-bc-ca)^{2}}{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+(ab+bc+ca)(a^{2}+b^{2}+c^{2})-2abc(a+b+c)}+3 \ge \frac{3(a+b+c)^{2}}{2(ab+bc+ca)}\]
which is not hard. :)

@note that the inequality holds for $ a=b=c$ and $ a=b,c=0.$ :)
[/hide][/quote]

After setting $ p=a+b+c$ , $ q=ab+bc+ca$ , $ r=abc$ I have obtained

$ 12p^{3}r+44q^{3}+5p^{4}q \geq 31p^{2}q^{2}+24pqr$

now I used SHUR with $ p=1$ that is $ 1+9r \geq 4q$ and $ 1 \geq 3q$ to obtain

$ 396q^{3}-279q^{2}+61q-4 \geq 0$ with $ 3q \le 1$

After decomposition

$ (3q-1)3(132q^{2}-147q+12) \geq 0$

which IS NOT TRUE!!!! :(

To see better the expression I rewrote it in explicit form as

$ 5\sum a^{5}(b+c)+12\sum a^{3}b^{3}+6abc\sum ab(a+b) \geq 5abc\sum a^{3}+21a^{2}b^{2}c^{2}+11\sum a^{2}b^{2}(a^{2}+b^{2})$

but also this seems FALSE !!!!

My conclusion is that this hint is not ok, but I can be wrong!!

Any help is very grateful.

Dear Ductrung,

can you please post your solution when you are not any more busy.

Thank you very much for your help. :oops:

[quote="manlio"]I like too much this inequality  :)  :wink:

I tried to solve it using

[quote="toanhocmuonmau"][quote="ductrung"]For any positive real numbers $ a,b$ and $ c$,
\[ \frac{1}{a^{2}+bc}+\frac{1}{b^{2}+ca}+\frac{1}{c^{2}+ab}\ge\frac{3(a+b+c)^{2}}{2(a^{2}+b^{2}+c^{2})(ab+bc+ca)}\]
[/quote]
[hide="hint"]
The inequality is equivalent to
\[ \sum\frac{b^{2}+c^{2}-bc}{a^{2}+bc}+3 \ge \frac{3(a+b+c)^{2}}{2(ab+bc+ca)}\]
By Cauchy Schwarz inequality, we have
\[ \sum\frac{b^{2}+c^{2}-bc}{a^{2}+bc}\ge \frac{(2(a^{2}+b^{2}+c^{2})-ab-bc-ca)^{2}}{\sum (a^{2}+bc)(b^{2}+c^{2}-ca)}=\frac{(2(a^{2}+b^{2}+c^{2})-ab-bc-ca)^{2}}{(ab+bc+ca)(a^{2}+b^{2}+c^{2})-4abc(a+b+c)}\]
Hence, it suffices to prove that
\[ \frac{(2(a^{2}+b^{2}+c^{2})-ab-bc-ca)^{2}}{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+(ab+bc+ca)(a^{2}+b^{2}+c^{2})-2abc(a+b+c)}+3 \ge \frac{3(a+b+c)^{2}}{2(ab+bc+ca)}\]
which is not hard. :)

@note that the inequality holds for $ a=b=c$ and $ a=b,c=0.$ :)
[/hide][/quote]

After setting $ p=a+b+c$ , $ q=ab+bc+ca$ , $ r=abc$ I have obtained

$ 12p^{3}r+44q^{3}+5p^{4}q \geq 31p^{2}q^{2}+24pqr$

now I used SHUR with $ p=1$ that is $ 1+9r \geq 4q$ and $ 1 \geq 3q$ to obtain

$ 396q^{3}-279q^{2}+61q-4 \geq 0$ with $ 3q \le 1$

After decomposition

$ (3q-1)3(132q^{2}-147q+12) \geq 0$

which IS NOT TRUE!!!! :(

[/quote]

Please stop at this step and using just $ 1+9r \geq 4q$ (then you replace $ r$ by $ \frac{1+4q}{9}$). Dont replace anymore $ q=\frac{1}{3}$. I think you may make mistake in computing, since the polynomial of $ q$ should have two roots $ q=\frac{1}{3}$ and $ \frac{1}{4}.$

a solution can be found in the doc attached here
[u]http://www.mathlinks.ro/Forum/viewtopic.php?t=116853[/u]

[quote="hungkhtn"][/quote][quote="manlio"]I like too much this inequality  :)  :wink:

I tried to solve it using

[quote="toanhocmuonmau"][quote="ductrung"]For any positive real numbers $ a,b$ and $ c$,
\[ \frac{1}{a^{2}+bc}+\frac{1}{b^{2}+ca}+\frac{1}{c^{2}+ab}\ge\frac{3(a+b+c)^{2}}{2(a^{2}+b^{2}+c^{2})(ab+bc+ca)}\]
[/quote]
[hide="hint"]
The inequality is equivalent to
\[ \sum\frac{b^{2}+c^{2}-bc}{a^{2}+bc}+3 \ge \frac{3(a+b+c)^{2}}{2(ab+bc+ca)}\]
By Cauchy Schwarz inequality, we have
\[ \sum\frac{b^{2}+c^{2}-bc}{a^{2}+bc}\ge \frac{(2(a^{2}+b^{2}+c^{2})-ab-bc-ca)^{2}}{\sum (a^{2}+bc)(b^{2}+c^{2}-ca)}=\frac{(2(a^{2}+b^{2}+c^{2})-ab-bc-ca)^{2}}{(ab+bc+ca)(a^{2}+b^{2}+c^{2})-4abc(a+b+c)}\]
Hence, it suffices to prove that
\[ \frac{(2(a^{2}+b^{2}+c^{2})-ab-bc-ca)^{2}}{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+(ab+bc+ca)(a^{2}+b^{2}+c^{2})-2abc(a+b+c)}+3 \ge \frac{3(a+b+c)^{2}}{2(ab+bc+ca)}\]
which is not hard. :)

@note that the inequality holds for $ a=b=c$ and $ a=b,c=0.$ :)
[/hide][/quote]

After setting $ p=a+b+c$ , $ q=ab+bc+ca$ , $ r=abc$ I have obtained

$ 12p^{3}r+44q^{3}+5p^{4}q \geq 31p^{2}q^{2}+24pqr$

now I used SHUR with $ p=1$ that is $ 1+9r \geq 4q$ and $ 1 \geq 3q$ to obtain

$ 396q^{3}-279q^{2}+61q-4 \geq 0$ with $ 3q \le 1$

After decomposition

$ (3q-1)3(132q^{2}-147q+12) \geq 0$

which IS NOT TRUE!!!! :(

[/quote][quote="hungkhtn"]

Please stop at this step and using just $ 1+9r \geq 4q$ (then you replace $ r$ by $ \frac{1+4q}{9}$). Dont replace anymore $ q=\frac{1}{3}$. I think you may make mistake in computing, since the polynomial of $ q$ should have two roots $ q=\frac{1}{3}$ and $ \frac{1}{4}.$[/quote]

Dear hungkhtn, I followed your hint but I think it doesn't work  :( if I am not wrong.

We have after setting $ p=a+b+c$ , $ q=ab+bc+ca$ , $ r=abc$

$ 12p^{3}r+44q^{3}+5p^{4}q \geq 31p^{2}q^{2}+24pqr$

Now we put $ p=1$ so it becomes

$ 12r+44q^{3}+5q \geq 31q^{2}+24qr$

or

$ 12r(1-2q)+44q^{3}+5q-31q^{2}\geq 0$

As $ (1-2q) \geq 0$ being $ 3q \le 1$ we have by Shur $ 1+9r \geq 4q$ or

$ r \geq \frac{4q-1}{9}$

Finally we are left with

$ 132q^{3}-125q^{2}+39q-4 \geq 0$

equivalent to

$ (q-\frac{1}{4})(q-\frac{1}{3})(132q-48) \geq 0$

which is FALSE. Am I wrong? :maybe:

Thank you very much.

[quote="manlio"][/quote][quote="hungkhtn"][/quote][quote="manlio"]I like too much this inequality  :)  :wink:

I tried to solve it using

[quote="toanhocmuonmau"][quote="ductrung"]For any positive real numbers $ a,b$ and $ c$,
\[ \frac{1}{a^{2}+bc}+\frac{1}{b^{2}+ca}+\frac{1}{c^{2}+ab}\ge\frac{3(a+b+c)^{2}}{2(a^{2}+b^{2}+c^{2})(ab+bc+ca)}\]
[/quote]
[hide="hint"]
The inequality is equivalent to
\[ \sum\frac{b^{2}+c^{2}-bc}{a^{2}+bc}+3 \ge \frac{3(a+b+c)^{2}}{2(ab+bc+ca)}\]
By Cauchy Schwarz inequality, we have
\[ \sum\frac{b^{2}+c^{2}-bc}{a^{2}+bc}\ge \frac{(2(a^{2}+b^{2}+c^{2})-ab-bc-ca)^{2}}{\sum (a^{2}+bc)(b^{2}+c^{2}-ca)}=\frac{(2(a^{2}+b^{2}+c^{2})-ab-bc-ca)^{2}}{(ab+bc+ca)(a^{2}+b^{2}+c^{2})-4abc(a+b+c)}\]
Hence, it suffices to prove that
\[ \frac{(2(a^{2}+b^{2}+c^{2})-ab-bc-ca)^{2}}{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+(ab+bc+ca)(a^{2}+b^{2}+c^{2})-2abc(a+b+c)}+3 \ge \frac{3(a+b+c)^{2}}{2(ab+bc+ca)}\]
which is not hard. :)

@note that the inequality holds for $ a=b=c$ and $ a=b,c=0.$ :)
[/hide][/quote]

After setting $ p=a+b+c$ , $ q=ab+bc+ca$ , $ r=abc$ I have obtained

$ 12p^{3}r+44q^{3}+5p^{4}q \geq 31p^{2}q^{2}+24pqr$

now I used SHUR with $ p=1$ that is $ 1+9r \geq 4q$ and $ 1 \geq 3q$ to obtain

$ 396q^{3}-279q^{2}+61q-4 \geq 0$ with $ 3q \le 1$

After decomposition

$ (3q-1)3(132q^{2}-147q+12) \geq 0$

which IS NOT TRUE!!!! :(

[/quote][quote="hungkhtn"]

Please stop at this step and using just $ 1+9r \geq 4q$ (then you replace $ r$ by $ \frac{1+4q}{9}$). Dont replace anymore $ q=\frac{1}{3}$. I think you may make mistake in computing, since the polynomial of $ q$ should have two roots $ q=\frac{1}{3}$ and $ \frac{1}{4}.$[/quote][quote="manlio"]

Dear hungkhtn, I followed your hint but I think it doesn't work  :( if I am not wrong.

We have after setting $ p=a+b+c$ , $ q=ab+bc+ca$ , $ r=abc$

$ 12p^{3}r+44q^{3}+5p^{4}q \geq 31p^{2}q^{2}+24pqr$

Now we put $ p=1$ so it becomes

$ 12r+44q^{3}+5q \geq 31q^{2}+24qr$

or

$ 12r(1-2q)+44q^{3}+5q-31q^{2}\geq 0$

As $ (1-2q) \geq 0$ being $ 3q \le 1$ we have by Shur $ 1+9r \geq 4q$ or

$ r \geq \frac{4q-1}{9}$

Finally we are left with

$ 132q^{3}-125q^{2}+39q-4 \geq 0$

equivalent to

$ (q-\frac{1}{4})(q-\frac{1}{3})(132q-48) \geq 0$

which is FALSE. Am I wrong? :maybe:

Thank you very much.[/quote]

Dear Manlio, that is what I mean, since you [b]are totally right[/b] ! Notice that if $ q\ge \frac{1}{4}$ then
\[ (q-\frac{1}{4})(q-\frac{1}{3})(132q-48)=48(q-\frac{1}{4})(q-\frac{1}{3})(q-\frac{4}{11}) \ge 0 \]
Otherwise, if $ q\le \frac{1}{4}$ then
\[ 12r(1-2q)+44q^{3}+5q-31q^{2}\geq 0 \]
since (putting $ r=0$)
\[ 12r(1-2q)+44q^{3}+5q-31q^{2}\geq 44q^{3}+5q-31q^{2}=44q(q-\frac{1}{4})(q-\frac{5}{11}) \ge 0, \]
That is all.

[quote="manlio"]Thank you very much, hungkhtn.  :)

You are really great  :wink:[/quote]

Just a small help for a close friend. Please and pleas dont say these too big word after  :(

[quote="can_hang2007"][quote="ductrung"][quote="manlio"]Thank you very much, hungkhtn and ductrung for your great help.  :)[/quote]
We are glad you're here with us, Manlio.[/quote]
Up to now, we dont have a simple solution for this inequality, so I will post the following solution (my second solution)-which I think is the simplest :). See it at: http://canhang2007.wordpress.com/2008/12/10/inequality-pham-huu-duc/ :)[/quote]

Really great your solution, can_hang2007. :roll:

 Congratulations. :)

[quote="ductrung"]For any positive real numbers $ a,b$ and $ c$,
\[ \frac {1}{a^{2} \plus{} bc} \plus{} \frac {1}{b^{2} \plus{} ca} \plus{} \frac {1}{c^{2} \plus{} ab}\ge\frac {3(a \plus{} b \plus{} c)^{2}}{2(a^{2} \plus{} b^{2} \plus{} c^{2})(ab \plus{} bc \plus{} ca)}\]
[/quote]
Let $ a\plus{}b\plus{}c\equal{}3u,$ $ ab\plus{}ac\plus{}bc\equal{}3v^2$ and $ abc\equal{}w^3.$
Hence, your inequality is equivalent to $ Aw^6\plus{}Bw^3\plus{}C\leq0,$
where $ A\equal{}\frac{4u^2}{(3u^2\minus{}2v^2)v^2}>0,$ $ B\equal{}B(u,v^2)$ and $ C\equal{}C(u,v^2).$
Thus,  $ Aw^6 \plus{} Bw^3 \plus{} C$ gets the maximal value on bounders of $ w^3.$
But $ w^3\equal{}x^3\minus{}3ux^2\plus{}3v^2x$ and  $ a,$ $ b$ and $ c$ are roots of the equation.
Id est, we obtain the bounders of $ w^3,$ when two of $ a,$ $ b$  and $ c$ are equal or when $ w^3\rightarrow0^\plus{}.$
Hence, we need check your inequality for $ b\equal{}c\equal{}1$ and for $ c\rightarrow0^\plus{},$ which is true.

We have also

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

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

[quote=ductrung]For any positive real numbers $a,b$ and $c$, \[\frac{1}{a^{2}+bc}+\frac{1}{b^{2}+ca}+\frac{1}{c^{2}+ab}\ge\frac{3(a+b+c)^{2}}{2(a^{2}+b^{2}+c^{2})(ab+bc+ca)}\][/quote]

[img]http://s12.sinaimg.cn/middle/006ptkjAzy76xBJKvaz0b&690[/img]
[img]http://s4.sinaimg.cn/middle/006ptkjAzy76xBLDx8ne3&690[/img]

Dear luofangxiang,

your inequality

$\sum \frac{b^2+c^2}{a^2+bc} \geq 2\sum \frac{a}{b+c} $ is very interesting, but I don't understand why

$\sum a^2 +\frac{6abc}{\sum a} + \frac{abc\sum a}{\sum a^2} \geq 2\sum ab $ is SHUR

and why

$ \sum a^2 + \frac{6abc}{\sum a} + \frac{abc\sum a} {\sum a^2} \geq 2\sum ab$ implies

 $ (\sum a^2)^2 + 4abc\sum \frac{a^2}{b+c} + \frac{abc\sum a} {\sum a^2} \geq 2\sum ab \sum a^2$

I don't remember if   $\sum \frac{a^2}{b+c} \geq \frac{3\sum a^2}{2\sum a} $ is true or false.

Thank you very much.

Dear luofangxiang,

now I understand your proof.

It is very nice, congratulations !!

$M={\frac {2}{27}}\, \left(  \left( {x}^{2}+yz \right) ^{-1}+ \left( {y
}^{2}+zx \right) ^{-1}+ \left( {z}^{2}+xy \right) ^{-1} \right)
 \left( {x}^{2}+{y}^{2}+{z}^{2} \right)  \left( xy+zx+yz \right)
 \left( x+y+z \right)
$

\[1/9\, \left( {\frac {yz}{x}}+{\frac { \left( z+x \right)  \left( x+y
 \right) }{y+z}} \right)  \left( {\frac {zx}{y}}+{\frac { \left( x+y
 \right)  \left( y+z \right) }{z+x}} \right)  \left( {\frac {xy}{z}}+{
\frac { \left( y+z \right)  \left( z+x \right) }{x+y}} \right)\geq M \geq 1/9
\, \left( x+y+z \right) ^{3}\]

\[1/9\, \left( {\frac {yz}{x}}+{\frac { \left( z+x \right)  \left( x+y
 \right) }{y+z}} \right)  \left( {\frac {zx}{y}}+{\frac { \left( x+y
 \right)  \left( y+z \right) }{z+x}} \right)  \left( {\frac {xy}{z}}+{
\frac { \left( y+z \right)  \left( z+x \right) }{x+y}} \right)\geq M \geq 1/16
\,{\frac { \left( y+z \right) ^{3} \left( x+y \right)  \left( z+x
 \right) }{{x}^{2}+yz}}+1/16\,{\frac { \left( z+x \right) ^{3} \left(
x+y \right)  \left( y+z \right) }{{y}^{2}+zx}}+1/16\,{\frac { \left( x
+y \right) ^{3} \left( y+z \right)  \left( z+x \right) }{{z}^{2}+xy}}\]

\[{\frac {1}{72}}\, \left( {\frac { \left( z+x \right)  \left( x+y
 \right) }{x}}+{\frac { \left( z+x \right)  \left( x+y \right) }{y+z}}
 \right)  \left( {\frac { \left( x+y \right)  \left( y+z \right) }{y}}
+{\frac { \left( x+y \right)  \left( y+z \right) }{z+x}} \right)
 \left( {\frac { \left( y+z \right)  \left( z+x \right) }{z}}+{\frac {
 \left( y+z \right)  \left( z+x \right) }{x+y}} \right)\geq M \geq 1/9\,
 \left( x+y+z \right) ^{3}\]

\[{\frac {1}{72}}\, \left( {\frac { \left( z+x \right)  \left( x+y
 \right) }{x}}+{\frac { \left( z+x \right)  \left( x+y \right) }{y+z}}
 \right)  \left( {\frac { \left( x+y \right)  \left( y+z \right) }{y}}
+{\frac { \left( x+y \right)  \left( y+z \right) }{z+x}} \right)
 \left( {\frac { \left( y+z \right)  \left( z+x \right) }{z}}+{\frac {
 \left( y+z \right)  \left( z+x \right) }{x+y}} \right)\geq M \geq 1/16\,{
\frac { \left( y+z \right) ^{3} \left( x+y \right)  \left( z+x
 \right) }{{x}^{2}+yz}}+1/16\,{\frac { \left( z+x \right) ^{3} \left(
x+y \right)  \left( y+z \right) }{{y}^{2}+zx}}+1/16\,{\frac { \left( x
+y \right) ^{3} \left( y+z \right)  \left( z+x \right) }{{z}^{2}+xy}}\]

\[1/8\,{\frac { \left( x+y \right) ^{2} \left( z+x \right) ^{2}}{y+z}}+
1/8\,{\frac { \left( x+y \right) ^{2} \left( y+z \right) ^{2}}{z+x}}+1
/8\,{\frac { \left( y+z \right) ^{2} \left( z+x \right) ^{2}}{x+y}}\geq M \geq
1/9\, \left( x+y+z \right) ^{3}\]

\[1/8\,{\frac { \left( x+y \right) ^{2} \left( z+x \right) ^{2}}{y+z}}+
1/8\,{\frac { \left( x+y \right) ^{2} \left( y+z \right) ^{2}}{z+x}}+1
/8\,{\frac { \left( y+z \right) ^{2} \left( z+x \right) ^{2}}{x+y}}\geq M \geq
1/16\,{\frac { \left( y+z \right) ^{3} \left( x+y \right)  \left( z+x
 \right) }{{x}^{2}+yz}}+1/16\,{\frac { \left( z+x \right) ^{3} \left(
x+y \right)  \left( y+z \right) }{{y}^{2}+zx}}+1/16\,{\frac { \left( x
+y \right) ^{3} \left( y+z \right)  \left( z+x \right) }{{z}^{2}+xy}}\]

\[1/4\,{\frac {{x}^{3} \left( y+z \right) ^{2}}{yz}}+1/4\,{\frac {{y}^{
3} \left( z+x \right) ^{2}}{zx}}+1/4\,{\frac {{z}^{3} \left( x+y
 \right) ^{2}}{xy}}\geq M \geq 1/9\, \left( x+y+z \right) ^{3}\]

\[1/4\,{\frac {{x}^{3} \left( y+z \right) ^{2}}{yz}}+1/4\,{\frac {{y}^{
3} \left( z+x \right) ^{2}}{zx}}+1/4\,{\frac {{z}^{3} \left( x+y
 \right) ^{2}}{xy}}\geq M \geq 1/16\,{\frac { \left( y+z \right) ^{3} \left( x+
y \right)  \left( z+x \right) }{{x}^{2}+yz}}+1/16\,{\frac { \left( z+x
 \right) ^{3} \left( x+y \right)  \left( y+z \right) }{{y}^{2}+zx}}+1/
16\,{\frac { \left( x+y \right) ^{3} \left( y+z \right)  \left( z+x
 \right) }{{z}^{2}+xy}}\]

\[1/4\,{\frac { \left( y+z \right) ^{2}yz}{x}}+1/4\,{\frac { \left( z+x
 \right) ^{2}xz}{y}}+1/4\,{\frac { \left( x+y \right) ^{2}xy}{z}}\geq M \geq 1/
9\, \left( x+y+z \right) ^{3}\]

\[1/4\,{\frac { \left( y+z \right) ^{2}yz}{x}}+1/4\,{\frac { \left( z+x
 \right) ^{2}xz}{y}}+1/4\,{\frac { \left( x+y \right) ^{2}xy}{z}}\geq M \geq 1/
16\,{\frac { \left( y+z \right) ^{3} \left( x+y \right)  \left( z+x
 \right) }{{x}^{2}+yz}}+1/16\,{\frac { \left( z+x \right) ^{3} \left(
x+y \right)  \left( y+z \right) }{{y}^{2}+zx}}+1/16\,{\frac { \left( x
+y \right) ^{3} \left( y+z \right)  \left( z+x \right) }{{z}^{2}+xy}}\]

\[1/4\,{\frac {{x}^{2} \left( x+y \right)  \left( y+z \right)  \left( z
+x \right) }{{y}^{2}+{z}^{2}}}+1/4\,{\frac {{y}^{2} \left( x+y
 \right)  \left( y+z \right)  \left( z+x \right) }{{x}^{2}+{z}^{2}}}+1
/4\,{\frac {{z}^{2} \left( x+y \right)  \left( y+z \right)  \left( z+x
 \right) }{{x}^{2}+{y}^{2}}}\geq M \geq 1/9\, \left( x+y+z \right) ^{3}\]

\[1/4\,{\frac {{x}^{2} \left( x+y \right)  \left( y+z \right)  \left( z
+x \right) }{{y}^{2}+{z}^{2}}}+1/4\,{\frac {{y}^{2} \left( x+y
 \right)  \left( y+z \right)  \left( z+x \right) }{{x}^{2}+{z}^{2}}}+1
/4\,{\frac {{z}^{2} \left( x+y \right)  \left( y+z \right)  \left( z+x
 \right) }{{x}^{2}+{y}^{2}}}\geq M \geq 1/16\,{\frac { \left( y+z \right) ^{3}
 \left( x+y \right)  \left( z+x \right) }{{x}^{2}+yz}}+1/16\,{\frac {
 \left( z+x \right) ^{3} \left( x+y \right)  \left( y+z \right) }{{y}^
{2}+zx}}+1/16\,{\frac { \left( x+y \right) ^{3} \left( y+z \right)
 \left( z+x \right) }{{z}^{2}+xy}}\]

[quote=ductrung]For any positive real numbers $a,b$ and $c$, \[\frac{1}{a^{2}+bc}+\frac{1}{b^{2}+ca}+\frac{1}{c^{2}+ab}\ge\frac{3(a+b+c)^{2}}{2(a^{2}+b^{2}+c^{2})(ab+bc+ca)}\][/quote]

[img]http://s16.sinaimg.cn/middle/006ptkjAzy77t3DVnGLbf&690[/img]


summer program
MathPath
AwesomeMath
AMC
USA(J)MO
USAJMO
USAMO

Hey guys, since its summer vacation what are you guys doing or going somewhere?

I just came back from Camp Tukabatchee for 1 week and going to MathPath for 4 weeks.
So, just post what you are doing during the summer... I feel isolated... :|

ok, it was all going bad until we played bbal a cuple times and thaan we went ice skating...(except for the part when patrick fell right in front of me and i hit my head on the ice :rotfl: )
it got better!

mines been pretty good. just got back from perry county for a mission trip and i played bball with gene and my other buddies and went ice skating:) ive got debate camp next week. then shocco for a church retreat then math camp. oh and today i played ultimate and tennis

ok. I've been working full time for 6 weeks, but today was my last day. Yay.

Now, hopefully I can average 12 hours of math a day for the next 5 weeks (but this seems unlikely, due to laziness, brain capacities, SAT prep, Summer Reading, etc. edit: oh yea and having a life).

and 3/2 hours of piano.

I'm going to AwesomeMath on Monday, and I'm gonna be gone for 3 weeks. Hmmm I should find out when my plane leaves....

my summer has been awesome, probably the best on record, and I'm brown from 2 hours of swim team practice a day  :D. I didn't go to debate camp or math camp or MASI or anything. I've got high school orientation tomorrow, and I'm starting freshman year Monday. Wow, no more MathCounts. Though I realized that on March 28th around 3:00 PM. It's okay, I still have AMC (who I am mad at because of this crazy new JMO. You heard me, AMC. I'm mad at you.) and local comps.

Will the Grissom teacher let you take the amc 12 as a freshman? If she will, you can get around the usajmo. However, just because the really smart middle schoolers/young high schoolers don't like the usajmo, it doesn't necessarily mean it's a bad thing. The vast majority (which is not really represented well by aopsers) of people will find taking the usajmo a better experience than the usamo, since most people prefer to not be forced to work 9 hours on 6 problems when they can't solve a single one. (less than half the people taking the usamo can solve 1 out of 6 questions on most years, and this is saying something, since the people taking the usamo are already a smart group of people).

However, if you can consistently solve one or 2 usamo problems on practice usamos, then you would be better off taking the usamo. old usamos can be found at

[url]http://www.unl.edu/amc/e-exams/e8-usamo/archiveusamo.shtml[/url]

It should be noted that old usamo number one's are often much easier than usamo number one's nowadays.

There are some rules concerning USAJMO that should be refined a bit, I agree, especially about the qualifications for red mop, but overall it's probably a good thing.

i love high school. it's amazing. spain park is the best school in the world.

screw you:) haha jk jk. well schools been aight so far.. i dunno if im in geometry or alg II yet.. haha dynamo i might even do both:) well... math team atm's kinda lame since we arn't doing anything.. fall startup in like 3 weeks:) anyone else doing it? besides gene and me?

I think I'm doing it. It was one of the more interesting in school tests last year.

Of course, this year for me, I think the more interesting one for me will be the one that involves calculus. I should find time to study calc...

I heard that Vestavia is going to HMMT. Are you going tony,gene? Mrs. Campbell said Eugene and I could go with Vestavia on the trip. Hope it doesn't get canceled like that other year.

I'm doing fall startup. :D  My goal is to get 75 answered and get 90% or higher accuracy... haha. I'll probably end up getting, like, a 50.

hey dynamo.. thats pretty lofty:) haha i got a 71 on the 06 one but i spent like 45 minutes on it.. anyways i still beat all the somphmores there:) and first place that year was 70 and 20th was around 40. :) yea hurdlers we are going there.. mr taylor never specified if freshman could go but he mentioned that there was a possibility that we could.. haha:)

yeah, it's very lofty, I know. But if I dont challenge myself how will I ever get better? I got 70 or so answered on the 07 one in 35 minutes but idk what I got on it. I need to learn to go faster.

Fall Startup for actually is this thursday!! Holy crap, my first high school competition kinda! well unless you count amc or arml, but i didnt actually go to grissom. Inhale, exhale, wheeze, wheeze. I am very nervous. I dont remember being this nervous for math team since our first tournament in 6th grade. I have to prove myself.

whats hmmt? Hmm? haha...

oh btw I took 2 amc's lately and i got a 132 and a 144, so it seems I havent really improved  :( . Aiming for a perfect on the 10 on first try, will get 126 then 138... what about you guys?

ok so this post has gone on forever, bye

to be honest, i haven't been doing that much math lately... :(

[quote="dynamo729"]
Fall Startup for actually is this thursday!! Holy crap, [/quote]

oh hmmm i just realized that hmmm I need to play some ftw :lol:

tony are you going to duke? - Eugene and i are.

Nick, I had heard Mrs. Campbell say something about harvard-mit(HMMT) and possibly duke take ppl from grissom to go as well, have you heard anything about this?

Ohhhh. HMMT = Harvard/MIT  :D
Yeah, I've heard of Harvard/MIT. Only upperclassmen are going (Jr/Sr). I think there was something like 7 people? I dunno, that's how many I counted on the board. I assume all of you guys are going :P. I'm not. D:

hurdler I love your ftw idea :D

actually I'll probably be sick by Thursday. All three of my siblings have swine flu, I'm hanging on... haha. or i might just go to school for the fall startup. and fail it.

hey people :D
i took the 2008 fall startup today and I got a 70. That would've been 5th last year... I need to improve :(. Haha before tomorrow.
oh, and our graduation exam for math is happening almost simultaneously! so I'm going to go super fast on that as warmup :D.

wow that is [u]SUCH a legit[/u] idea:) im doing it then:) haha wow 70 on 2008? dang... :) haha yea grissom people are going to duke too but im not going to duke bc of stuff and since there's like 3 somphmores going and no freshies (one if i go)

nice, a 70... i think i had a 72 on that one... and probably like a forty something in 9th grade cuz i triple checked everything for some reaon

hmmm i double checked for some reason way too much. I guess i have been spending too much on problems that require like 2 hours each (at least for me cuz i try to do them while half listening to a teacher in school). oh well. Next year I am going to start from the back page cuz the back page had ez prbs i missed. hopefully i will also stop doubting my addition skills if i start off with harder probs

wow... um, the official contest was over yesterday, so I'll tell you I got a LOT less on this years :( . I despise nerves and switching the answers for problems :(

yea my friend said he was off on a lot of numbers apparently (he answered 59 and then realized it was the 69 box after the contest) :(

i am guessing i did worse than last year cuz last year i didn't spend time double checking and actually got to number 100 (as a result of skipping tons of probs that i guessed would take me 30 secs+) (not that i could do 100 since i didn't know any calc then) this year i lost all speed test taking skills.

yea pretty much i skipped around and did everything that i could do in <30 seconds and i didn't do any double checking...

nice job i heard you beat me by 5 with a 73. :)

whoa i didnt know you were in woot too tony. i did it last year and it was pretty fun, except the part where i got lazy and didn't do most of the problems, (or was too busy)

woot is hard....

tony beat me by 21 on fall startup, bah :P. Xinke beat me by 16.

Ohh.. btw Xinke, are you the student who gave the quote at the bottom of this page? [url]http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=206[/url]

I TOOK A PRACTICE AMC10 AND GOT A 150 :D

...what? who cares that I've taken it before? It was too long ago to remember the answers, anyways.

PS Tony you are weird for taking woot..


I have been given at school some problems and this is one I can't manage to solve! :(

Perhaps someone will help! :D
Let f be a function, an incresing function, f is defiened on (0,+oo) -->(0,1). Lets consider now a sequence (a_n)n>=1 which has only pozitive values (a_n is in (0,+oo)) and it has the following property:
a_(n+1)=f(a_n)*max{ ((a_n+a_(n+1))/2, a_n)} for any n in N. Prove that:
1)(a_n) converges;
2) lim(n-->oo) a_n=
a)1/e; b) 1; c)oo; d) e; f)0; g)another value;
if g is the answer then what is the value of  the limit?

thank you very much! bye bye!

Well, part (a) isn't bad at all.  We basically just want to put a nice bound on the sequence a(n).  Try to do it recursively -- that is, what can you say about a(n+1) is less than or equal to, in terms of a(n)?

Don't use these hints until you've tried the problem:

Hint #1:

[hide]f(x) :le: 1[/hide]

Hint #2:

[hide]Look at the max funtion -- break it into two cases.[/hide]

Hint #3:

[hide]Any monotonic bounded sequence of real numbers has a limit.[/hide]

I find for the second part that the limit isn't necessarily fixed -- I think I can choose f to make it converge to any non-negative real number.


calculus integration geometry trigonometry blogs

The (net) area under the graph of the equation you're integrating in $[0,\pi]$. Just read a calculus textbook; you'll learn more than reading this.  ;)

Masoud Zargar

[quote="lotrgreengrapes7926"]Do you want a definition of integral? Or an answer to the question? The integral is the area bound by the graph $\frac{x\sin{x}}{1+\cos^2{x}}$ and the x-axis between 0 and $\pi$.[/quote]
 Just the definition of what it is and how it works. If you want you can solve it for fun

[quote="Sir.Aaron"]Just the definition of what it is and how it works. If you want you can solve it for fun[/quote]
I don't know how.  :blush:

I don't know if this will help, but $\displaystyle\int^{\pi}_0 x\sin{x}\,dx = \pi$ and $\displaystyle\int^{\pi}_0 1+\cos^2{x}\,dx = 1.5\pi$. I figured that out with a graphing calculator. :D

[quote="JBL"]You really are going to have to go learn calculus if you want that.  [url=http://mathworld.wolfram.com/RiemannIntegral.html]Mathworld[/url] has lots of information, not all of which will be easy to decipher.[/quote]
Ok,I'll do my reading.

[quote="deej21"]That doesn't help... you can't separate it like that.  You'll have to use some integration techniques... although I don't think substitution or integration by parts will work here.[/quote]

Look at the site from which the image is taken, you'll find a nice solution using symmetry.  :)

[quote="paladin8"][quote="Sir.Aaron"]What does intagration mean? Like in this example:
[img]http://www.wangsblog.com/jeffrey//pictures/9dee518d387fa0140071b2d6b6849f7f.gif[/img][/quote]

WHOA this looks really familiar maybe it's from my blog! ;)[/quote]
Sure is. Very nice place by the way. I hope you dont mind but I put you blog link on [url=http://crypt3x.blogspot.com/]MY BLOG[/url]


function
calculus
derivative
calculus computations

I have a function $0.5\sqrt{L^{4}}\sqrt{K}$ and I have to find part elasticity for L and K.

Elasticity for an unknown (dunno if they teach elasticity to everyone) is $E_{L}f=\frac{L}{f(K, L)}* f'_{L}(K, L)$

The problem is, I did it by hand and arrived to the conclusion that both answers are 0.5.

When I went to check the answers with Mathematica, it reported them as $0.5KL^{4}$ for K and $2KL^{4}$ for L.

Have I made a mistake putting the formulas into Mathematica or do I just plain old suck at maths  :oops: ?

Afternote: might have posted this in the wrong forum, is this forum for easy unsolved calculus problems aswell or should I post these things into the other forum from now on if they should occur?

They do teach elasticity to (almost) everyone, but it is called a logarithmic derivative.  :)

Simplify the function to $0.5L^{2}K^{1/2}$ before differentiating. Taking a partial derivative, you get a constant multiple equal to the relevant exponent (either $2$ or $1/2$); the exponent is reduced by one, but then restored by multiplication. So, the answer is $2$ for $L$ and $0.5$ for $K$.

Cheers.

A logarithmic derivative  :o I'm glad I'm a business student, we seem to have easier names for things  :lol:


geometry
LaTeX
number theory
prime factorization

How do you make a header be more than one line long?
I looked here:
http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuidePack.php#fancy
When I put lots of text in the footer and the header, the different sections overlapped.

Also, after hitting ctrl and F7, and then pressing F5 to see it, I cannot see newer changes to it after repeating this.

You'll need to make more room for the header and footer. The best way to do this is by using the [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/geometry/manual.pdf]geometry package[/url]. You could decrease the top and bottom margins
\usepackage[top=[i]length[/i],bottom=[i]length[/i]]{geometry}
or else reduce the total body height
\usepackage[textheight=[i]length[/i]]{geometry}
or both. The geometry package takes care of all the other variables. Full details are in the [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/geometry/manual.pdf]manual[/url].

How do you find out your errors after building?
Can you use more than one package at a time? How?

If you wanted to have a running header with 5 lines, how would you set it up?

Let's say you wanted to use the AMS package (which comes standard with most Latex installations I believe). Simply include the commands:

[code]
\usepackage{amsmath,amssymb}
[/code]

And then you can use all the special commands that the AMS package provides.

As far as I know, I build my document every once in a while so that I can track down errors better. I am pretty sure it is a bad idea to only build a document only when you are finished it because a missing bracket can sometimes cause 7-100 errors alone  :(

Not sure about the header question  :oops:

[quote="aufha"]How do you find out your errors after building?[/quote]What are you using to write your tex document? If it's TeXniCenter then the errors are in the bottom window. Otherwise look in the log file that is created.
[quote="aufha"]Can you use more than one package at a time? How?[/quote]Either put them together as Prime factorization says or use lots of lines with
\usepackage{[i]package name[/i]}
[quote="aufha"]If you wanted to have a running header with 5 lines, how would you set it up?[/quote]Set what up? For the margins and text height just use trial and error until the page looks correct.

where is the log file? I don't see the errors in the bottom window.


inequalities
linear algebra
matrix
inequalities proposed

[quote="pohoatza"]Let $x_{i}$ real numbers, with $i \in \overline{1,n}$. Prove that: $\frac{\sum{\left( x_{i}-\left(\frac{\sum{x_{i}}}{n}\right)\right)^{2}}}{n}\leq \left(\frac{\max{x_{i}}-\min{x_{i}}}{2}\right)^{2}$[/quote]
[b]Notation:[/b]
$X_{n}: =\{x_{1},x_{2},...,x_{n}\}\; ,\; X_{n}\subset{\mathbb R}$
the symbol   $[\cdot]$  denotes the integer part,
 $\bar{\mathbf x}: = \frac{1}{n}\sum_{k=1}^{n}x_{k}\; ,$
 ${\mathbf \Delta}: = n\sum_{k=1}^{n}\left(x_{k}-\bar{\mathbf x}\right)^{2}=\sum_{1\le i\le j\le n}(x_{i}-x_{j})^{2}=n\sum_{k=1}^{n}x_{k}^{2}-\left(\sum_{k=1}^{n}x_{k}\right)^{2}$
${\mathbf d}: = \max_{1\le i\le n}{x_{i}}-\min_{1\le i\le n}{x_{i}}$

Using these notation, the interesting question proposed by Cosmin is to prove $\begin{array}{|c|}\hline \\ \frac{2}{n}\sqrt{\mathbf \Delta}\le{\mathbf d}\\ \\ \hline \end{array}\; .$ A bit more precise result is :
[u][b]Proposition.[/b][/u] [i] We have $(*)\; \; \; \; \begin{array}{|c|}\hline \\ \sqrt{\frac{{\mathbf \Delta}}{\left[\frac{n}{2}\right]\left[\frac{n+1}{2}\right]}}\le{\mathbf d}\le \sqrt{\frac{2}{n}{\mathbf \Delta}}\\ \\ \hline \end{array}\; .$
 At the left  in (*) the equality holds iff $X_{n}$ are the roots of  $f_{*}(x)=\left(x-\bar{\mathbf x}\mp \sqrt{\frac{\left[\frac{n+1}{2}\right]}{\left[\frac{n}{2}\right]}\Delta }\right)^{\left[\frac{n}{2}\right]}\left(x-\bar{\mathbf x}\pm\frac{1}{n}\sqrt{\frac{\left[\frac{n}{2}\right]}{\left[\frac{n+1}{2}\right]}{\mathbf \Delta}}\right)^{\left[\frac{n+1}{2}\right]}\; .$
  At the right, the equality case is valid iff $X_{n}$ are the roots of  $f^{*},$ where [/i]  $f^{*}(x)=\left((x-\bar{\mathbf x})^{2}-\frac{{\mathbf \Delta}}{2n}\right) (x-\bar{\mathbf x})^{n-2}.$

[u][b]Proof:[/b][/u] Assume $x_{1}\ge x_{2}\ge ...\ge x_{n}$ , so that ${\mathbf d}=x_{1}-x_{n}\; .$  Let us prove the inequality  ${\mathbf d}\le \sqrt{\frac{2}{n}{\mathbf \Delta}}\; \; \; .$ We have
\[{\mathbf \Delta}=n(x_{1}-x_{n})^{2}+2n(\bar{\mathbf x }-x_{1})(\bar{\mathbf x }-x_{n})+n\sum_{k=2}^{n-1}(x_{k}-\bar{\mathbf x })^{2}\; \; \; . \]
Observe that   $( \bar{\mathbf x }-x_{1})(\bar{\mathbf x }-x_{n})\ge-\frac{{\mathbf d}^{2}}{4}$ with equality iff   $\bar{\bf x }=\frac{1}{2}(x_{1}+x_{n}) .$ But this implies
\[{\mathbf \Delta}\ge \frac{n}{2}{{\mathbf d}^{2}}\; \; \; , \; \; \; \mbox{that is}\; \; \;{\mathbf d}\le \sqrt{\frac{2}{n}{\mathbf \Delta}}\; . \]
If $X_{n}$ are the roots of $f(x),$ then $f(x)$   must have the form $f(x)=(x-x_{1})(x-x_{n})(x-\bar{\mathbf x })^{n-2}\; \; \; ,$ that is $f=f^{*}\; .$

[b] The lower bound may[/b] be obtained in following way: from ${\mathbf \Delta}=\sum_{1\le i\le j\le n}(x_{i}-x_{j})^{2}$   we find  ${\partial{\mathbf \Delta}\over\partial x_{i}}=2n(x_{i}-\bar{\mathbf x }) \; \; \; .$ By assuming $x_{1}\ge \, ... \ge x_{k}\ge \bar{\mathbf x }\ge x_{k+1}\ge \, ... \ge x_{n}\; ,$ it's easy to obtain  ${\mathbf \Delta}\le k(n-k)(x_{1}-x_{n})^{2}\; \; \; .$
On the other hand  $k(n-k)\le \left[\frac{n}{2}\right]\cdot \left[\frac{n+1}{2}\right]\; ,$ with equality iff   $k=\left[\frac{n}{2}\right] \; \; \; \mbox{or}\; \; \; k=\left[\frac{n+1}{2}\right] \; .$
Therefore $X_{n}$ must be the set of roots of  $(x-x_{1})^{k}(x-x_{n})^{n-k}\; \; \; ,\quad k=\left[\frac{n}{2}\right] \; \; \; \textrm{sau}\; \; \; k=\left[\frac{n+1}{2}\right]\; ,$
i.e. the roots of  $f_{*}$ . In conclusion ${\mathbf d}\ge \sqrt{\frac{{\mathbf \Delta}}{\left[\frac{n}{2}\right]\cdot\left[\frac{n+1}{2}\right]}}\; .$

[u][b] Another method to find the lower bound in (*)[/b][/u]   is to use a discrete variant of the so-called "Gruss Inequality"  (see D.S.Mitrinovic "Analytic Inequalities",Springer Verlag,1970,  3.2.26 p.205) :
  [i] If  $a\leq t_{i}\leq A, \, b\leq z_{i}\leq B (i\in \{1, \, ..., n\}) ,$ then
\[\left|n\sum_{k=1}^{n}t_{k}z_{k}-\sum_{k=1}^{n}t_{k}\cdot \sum_{k=1}^{n}z_{k}\right|\le (A-a)(B-b)\left[\frac{n}{2}\right]\cdot\left[\frac{n+1}{2}\right] \]
with equality for
\[t_{1}=t_{2}= \, ... =t_{\left[\frac{n}{2}\right]}=t, \quad t_{\left[\frac{n+1}{2}\right]}= \, ... = t_{n}=T \; \; \; , \]

\[z_{1}=z_{2}= \, ... =z_{\left[\frac{n}{2}\right]}=z \; \; \; , \; \; \; z_{\left[\frac{n+1}{2}\right]}= \, ... = z_{n}=Z \; \; \; , \]
where the matrix  $\left\| \begin{array}{ll}t & T\\ z & Z \end{array}\right\|$ is one of the following $\left\| \begin{array}{ll}a & A\\ b & B \end{array}\right\| \; \; \; , \; \; \; \left\| \begin{array}{ll}A & a\\ B & b \end{array}\right\| \; \; \; , \; \; \; \left\| \begin{array}{ll}a & A\\ B & b \end{array}\right\| \; \; \; , \; \; \; \left\| \begin{array}{ll}A & a\\ b & B \end{array}\right\|\; .$ [/i]
If we select $t_{k}=z_{k}=x_{k}-\bar{\mathbf x }\; \; \; , \; \; \; A=B=x_{1}-\bar{\mathbf x }\; \; \; , \; \; \; a=b=x_{n}-\bar{\mathbf x }\; ,$
it follows $n\sum_{k=1}^{n}(x_{k}-\bar{\mathbf x })^{2}\le \left[\frac{n}{2}\right] \cdot\left[\frac{n+1}{2}\right]{\mathbf d}^{2}\; \; \; , \; \; \; \textrm{i.e.}\; \; \;{\mathbf d}\ge \sqrt{\frac{{\mathbf \Delta}}{\left[\frac{n}{2}\right]\left[\frac{n+1}{2}\right]}}\; .$

very nice mr. Lupas, i'm glad you showed your interest in this problem.

[quote="pohoatza"]Let $x_{i}$ real numbers, with $i \in \overline{1,n}$. Prove that:
$\frac{\sum{( x_{i}-(\frac{\sum{x_{i}}}{n}))^{2}}}{n}\leq  (\frac{\max{x_{i}}-\min{x_{i}}}{2})^{2}$[/quote]
With $f : \mathbb{R}^n \to \mathbb{R}$ given by:
\[ f(x_1,x_2,\ldots,x_n) = n^2(\max\{x_i\} - \min\{x_i\})^2 - 4n\sum_{i=1}^n x_i^2 + 4\left ( \sum_{i=1}^n x_i \right )^2 \]
We should prove that $f$ is non-negative. It is enough to prove that $f(x) \ge 0$ for all $x \in C$, where $C \subseteq \mathbb{R}^n$ is compact, since $\mathbb{R}^n = \bigcup_{k \in \mathbb{N}} [-k,k]^n$.
Let $C \subseteq \mathbb{R}^n$ be compact, and let $(x_1,\ldots,x_n) \in C$ be a minimum of $f$ on $C$. Wlog assume that $x_1 \le \cdots \le x_n$, since $f$ is symmetric. Assume for the sake of contradiction that there exists $k \in \{1,\ldots,n\}$ s.t. $x_k \notin \{x_1,\overline{x},x_n\}$, with $\overline{x} = \frac{1}{n} \sum_{i=1}^n x_i$. Then $0 = \frac{\partial f}{\partial x_k}(x_1,\ldots,x_n) = -8nx_k + 8\left ( \sum_{i=1}^n x_i \right ) = 8n(\overline{x}-x_k)$, so $x_k = \overline{x}$. Contradiction. Hence $x_k \in \{x_1,\overline{x},x_n\}$ for all $k$. Let $a=x_1,b=\overline{x},c=x_n$, and let $s,t$ be s.t. $x_1 = \cdots = x_s = a$ and $x_n = \cdots = x_{n-t+1} = c$. Then $b = \frac{as+ct}{s+t}$, and $x_{s+1} = \cdots = x_{n-t} = b$. And:
\[ \frac{1}{n}f(x_1,\ldots,x_n) = n(c-a)^2 - 4\left ( sa^2 + tc^2 + (n-s-t)b^2 \right ) + 4nb^2 \]\[
= n(c-a)^2 + 4(s+t)b^2 - 4sa^2 - 4tc^2 \ge (s+t)(c-a)^2 + 4(s+t)b^2 - 4sa^2 - 4tc^2 \]\[
= (s+t)(c-a)^2 + 4\frac{(sa+tc)^2}{s+t} - 4sa^2-4tc^2 = \frac{(s-t)^2(c-a)^2}{s+t} \ge 0 \]
Hence $f$ is non-negative on every compact interval of $\mathbb{R}^n$, and therefore it is non-negative on $\mathbb{R}^n$.


inequalities
function
inequalities theorems

I am just learning how to solve olympiad style inequalities, and to me mixing of variables is one of the most difficult and mysterious of methods. The concept is very simple: think of your inequality as some function $f$. You want to show that $f(x_{1}, ..., x_{n}) \ge S$, or something like that. So think of some simpler combination of variables, say $x_{1}, ..., x_{n-2}, \frac{x_{n-1}+x_{n}}{2}, \frac{x_{n-1}+x_{n}}{2}$, and show that $f(\text{original}) \ge f(\text{simpler})$, and then showing $f(\text{simpler}) \ge S$ may be easier.

The problem (for me at least) is this: How do you recognize when to use this method? And how do you know what "simpler" combination of variables to try? Of course experience is the best teacher. But all of the applications of mixing variables that I read on this forum were beyond me. So I went on a spree of searching for problems to which this method might be applied. I found a very simple one, and I thought it might be helpful for people like me to see this example.

Nesbitt's Inequality: For $a, b, c > 0$,
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$

I think Hojoo Lee gives 6 proofs of this inequality in his notes, but none of them use mixing variables. This method is huge overkill for an inequality that can be proven by AM-GM, but that's why it's a learning example. Anyway, first we will show that
$f(a, \frac{b+c}{2}, \frac{b+c}{2}) \le f(a, b, c)$
Then if we can show that $f(a, b, b) \ge \frac{3}{2}$, we win.

So we need to prove
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{a}{b+c}+\frac{b+c}{b+c+2a}+\frac{b+c}{b+c+2a}$

After canceling the common term and multiplying both sides by the denominators, it is

$b(b+a)(b+c+2a)+c(c+a)(b+c+2a) \ge 2(a+b)(a+c)(b+c)$

Upon expanding and canceling terms, this becomes

$b^{3}+ab^{2}+c^{3}+ac^{2}\ge 2abc+b^{2}c+bc^{2}$

which follows from three applications of AM-GM:

$\frac{b^{3}+b^{3}+c^{3}}{3}\ge b^{2}c$
$\frac{b^{3}+c^{3}+c^{3}}{3}\ge bc^{2}$
$\frac{ab^{2}+ac^{2}}{2}\ge 2abc$

Now all that's left is $f(a, b, b) \ge \frac{3}{2}$. But now there are only two variables, so it's a lot easier:

$\frac{a}{2b}+\frac{2b}{a+b}\ge \frac{3}{2}$

Clearing denominators, this is

$a(a+b)+(2b)^{2}\ge 3(b)(a+b)$
$a^{2}+ab+4b^{2}\ge 3ab+3b^{2}$
$(a-b)^{2}\ge 0$

which is trivially true.

We have equality when $a=b=c$.

Observe that this solution is much longer than most other proofs of Nesbitt's Inequality, but (aside from the application of mixing of variables) all of the steps are totally elementary and sort of present themselves, even to an inexperienced problem solver like myself.

Here are some threads (certainly not all!) on this forum which use mixing of variables. These are not as simple.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=variables+mixing&t=104630]arqady[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=variables+mixing&t=102482]phamduyhiep[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=variables+mixing&t=105408]Namdung[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=variables+mixing&t=101476]arqady[/url]

Experienced problem solvers, please post (or link to) problems that you think are well suited to mixing variables! Also discuss the method if you like.
I hope this helps! :D :D

http://www.mathlinks.ro/Forum/viewtopic.php?t=107851

This one is really difficult if someone tried to attack this with muirhead or even schur. However, it can be solved by mixing variables. I will post the solution later.

So, we can view the AM-GM inequality for two variables as a fundamental theorem of Mixing Variables  :D
The multivariable AM-GM inequality can be deduced from a(n in)finite applications of Mixing Variables Techniques.
Is it possible that we can prove the Cauchy-Schwarz inequality via Mixing Variables Techniques? (Think about the equality conditions...)

Wow, these are kinda cool, because it is really an elementary technique, not any kind of theorem that needed proving. Thanks for the post Xevarion. :)


Let $ f$ function such that :
$ (\forall x \in \mathbb{R} ): f(x^{2}\plus{}x\plus{}3)\plus{}2f(x^{2}\minus{}3x\plus{}5)\equal{}6x^{2}\minus{}10x\plus{}17$

Evaluate : $ f(85)$

Solve $ x^2\plus{}x\plus{}3\equal{}85$ to get $ x\equal{}\frac{1}{2}(\minus{}1\pm \sqrt{329})$, and $ x^2\minus{}3x\plus{}5\equal{}85$ to get $ x\equal{}\frac{1}{2}(3\pm\sqrt{329})$.

Now, note that if $ a\equal{}\frac{1}{2}(3\plus{}\sqrt{329}$ and $ b\equal{}\frac{1}{2}(\minus{}1\minus{} \sqrt{329})$, $ a^2\plus{}a\plus{}3\equal{}b^2\minus{}3b\plus{}5\equal{}c$.

Thus $ f(c)\plus{}2f(85)\equal{}6a^2\minus{}10a\plus{}17\equal{}509\plus{}4\sqrt{329}$ and $ f(85)\plus{}2f(c)\equal{}6b^2\minus{}10b\plus{}17\equal{}517\plus{}8\sqrt{329}$

Solving these equations we get $ f(85)\equal{}167$.

hello.  i think the answer is $ 171\frac{1}{3}$


inequalities
induction
function
inequalities open

Consider the following inequality: [tex]k(a^3+b^3+c^3-3abc)\geq (a+b+c)^3-27abc[/tex]. It is easy to show that for [tex]k\geq 4[/tex] the inequality holds for all a,b,c>0 and for [tex] k\leq 1[/tex] the reverse inequality holds.

Open question:what happens if  [tex]k\in(1,4)[/tex]?

Take [tex]c=0[/tex] (or [tex]c\rightarrow0[/tex]). If [tex]x=\sqrt{\frac a b}[/tex] then [tex]\frac{a^3+b^3}{(a+b)^3}=1-\frac3{(x+\frac1x)^2}[/tex]. It can take any value between 0.25 and 1.

Ok. Now let's think about the inquality [tex] k(a^4+b^4+c^4+d^4-4abcd)\geq (a+b+c+d)^4-4^4abcd [/tex]. What are the values of k in this case? ;)

Oh, I have a very ugly solution, which i don't know it it's correct,  for this one but since it is an open question  ;) , I will post it:
  Lemma (the most idiot result ever found-well, there are some others, friends know why):
  Observation: I really don't know if this is true or not, so I beg for someone interested to prove or to disprove it for a=b=1-2c.
  If $ a+b+c=1$ then we have $ 27(a^4+b^4+c^4)+\frac{148}{3}\cdot abc\geq\frac{229}{81} $.
   Proof:
   This is quite a strong inequality. Applying Cebasev and Schur wouldn't make a proof. So, again mixing variables. Let c the minimum among a,b,c. Then one can check that the inequality $ f(a,b,c)\geq f(\frac{a+b}{2},\frac{a+b}{2},c) $ is equivalent to $\frac{27}{8}\cdot [ 3(a+b)^2+4(a^2+ab+b^2)]\geq\frac{37}{3}\cdot c$. But this is true, since we have $ 4(a^2+ab+b^2)\geq 3(a+b)^2 $ and so it is enough to prove that $\frac{9}{2}\cdot (1-c)\geq\sqrt{\frac{37}{3}\cdot c} $
 which is true since $c\leq\frac{1}{3} $. Lemma is semi-proved. It remains to show that $ f(x,x,1-2x)\geq 0$ for all $0\leq x\leq\frac{1}{3} $, but this is really too ugly (even uglier than my solution) for me.
  So, hoping that the lemma holds, I will continue. I will prove that for k=27 the inequality is true. Let $ f(a,b,c,d) =LHS-RHS$ . Proving that $ f(a,b,c,d)\geq f(\frac{a+b+c}{3},\frac{a+b+c}{3},\frac{a+b+c}{3},d) $ reduces to $ 27\cdot[ a^4+b^4+c^4-(\frac{a+b+c}{3})^4]\geq 148\cdot d\cdot [(\frac{a+b+c}{3})^3-abc] $. But we can take d the minimum among the four numbers. Then $ d\leq\frac{a+b+c}{3} $ and we are left with proving the inequality $ 27\cdot[a^4+b^4+c^4-(\frac{a+b+c}{3})^4]\geq 148\cdot\frac{a+b+c}{3}\cdot[ (\frac{a+b+c}{3})^3-abc] $. Since this is homogenuous, we can take $ a+b+c=1$ and in this case we apply the lemma ( again, I'm really not sure that the lemma works!).
  Could anyone check my computations and the veridicity of the lemma? Note that 27 would be the best constant k since we could take a=b=c and d close to 0.

I've checked the computations and they are correct. So indeed [tex] k=27 [/tex] for the [tex]n=4[/tex] case. A far bigger challenge is to think about the general case. I can only assume that in that case [tex]k=(n-1)^{n-1}[/tex].

Maybe it's insane what I am about to say, but here goes nothing. We have to find [tex]\displaystyle\inf\frac{a_{1}^{n}+a_{2}^{n}+\ldots+a_{n}^{n}-na_1a_2\ldots a_n}{(a_1+a_2+\ldots+a_n)^n-n^na_1a_2\ldots a_n}[/tex]. Now I assume that this infimum is equal to [tex]\displaystyle\inf\frac{a_{1}^{n}+a_{2}^{n}+\ldots+a_{n-1}^{n}}{(a_1+a_2+\ldots+a_{n-1})^n}=(n-1)^{n-1}[/tex]. If we prove this last assertion, we solved the problem. :lol:

Wow, I don;t understand what you are saying, but the last assertion is not correct. I agree that we have to prove that $ k=(n-1)^{n-1} $ but I find this terribly difficult.

Ok, I'm too tired now to choeck for errors. I will prove by induction that for any $ a_1,...,a_n>0 $ with product 1 we have $ a_1^n+...+a_n^n-n\geq\frac{(a_1+...+a_n)^n-n^n}{(n-1)^{n-1}} $. Suppose it is true for n-1, let $ a_1,...,a_n>0 $ with product 1 and let the function $ f(a_1,...,a_n)=LHS-RHS$. Let
 t the geometric mean of $ a_1,...,a_{n-1} $ and I will prove that $ f(a_1,...,a_n)\geq f(t,t,...,t,a_n) $. Ok, this reduces to:
  $ a_1^n+...+a_{n-1}^n-(n-1)\cdot\sqrt[n-1]{a_1^n\cdot...\cdot a_{n-1}^n}\geq \frac{(a_1+...+a_n)^n-(a_n+(n-1)t)^n}{(n-1)^{n-1}} $. Now, let $ a_1+...+a_{n-1}=S, (n-1)t=P $. It is easy to prove using the bynomial formula and the fact that $ a_n\leq\sqrt[n-1]{a_1\cdot....\cdot a_{n-1}} $ ( I forgot to say that, I choose $ a_n$ the minimum among the n numbers) that we have the inequality $\frac{(a_1+...+a_n)^n-(a_n+(n-1)t)^n}{(n-1)^{n-1}}\leq\frac{(a_1+...+a_{n-1}+\sqrt[n-1]{a_1\cdot...\cdot a_{n-1}})^n-n^n\cdot\sqrt[n-1]{(a_1\cdot...\cdot a_{n-1})^n}} {(n-1)^{n-1}} $. Thus we have to prove that the moster from the RHS is at most $a_1^n+...+a_{n-1}^n-(n-1)\cdot\sqrt[n-1]{a_1^n\cdot...\cdot a_{n-1}^n}$. I will stop here.

Let's continue with proving the last inequality, for any positive reals $ a_1,...,a_{n-1} $. Since it is homogenuous, we may assume that $ a_1\cdot...\cdot a_{n-1}=1 $. Thus we have to prove that if $ a_1\cdot...\cdot a_{n-1}=1 $ then we have
  $ a_1^n+...+a_{n-1}^n-n+1\geq\frac{(1+a_1+...+a_{n-1})^n-n^n}{(n-1)^{n-1}} $. Now, I will use a deep and nice inequality created by Suranyi. The proof is left as exercise:
   For any positive reals $ x_1,...,x_n $ we have $ (n-1)(x_1^n+...+x_n^n)+n\cdot x_1\cdot...\cdot x_n\geq (x_1+...+x_n)\cdot (x_1^{n-1}+...+x_n^{n-1}) $.
  Now, take in that inequality $ x_i=a_i, i=1,n-1$ and $ x_n=1$. We find that
 $ a_1^n+...+a_{n-1}^n\geq\frac{(1+a_1+...+a_{n-1})(1+a_1^{n-1}+...+a_{n-1}^{n-1})-n}{ n-1}-1$. Also the inductive hypothesys shows that we also have the inequality $ a_1^{n-1}+...+a_{n-1}^{n-1}-n+1\geq\frac{(a_1+...+a_{n-1})^{n-1}-(n-1)^{n-1}}{(n-2)^{n-2}} $. Thus we can denote $ x=a_1+...+a_{n-1}\geq n-1$ and we have that $ a_1^n+...+a_{n-1}^n\geq\frac{(1+x)(n+\frac{x^{n-1}-(n-1)^{n-1}}{(n-2)^{n-2}})-n}{n-1}-1$. I will post again since Internet is very nasty tonight.

And now the most delicate part! We have to prove a monster . This monster would make jealous even Lagrangia:
   For any $ x\geq n-1$ we have the inequality
   $\frac{(1+x)(n+\frac{x^{n-1}-(n-1)^{n-1}}{(n-2)^{n-2}})-n}{n-1}\geq n+\frac{(x+1)^n-n^n}{(n-1)^{n-1}} $. I think this can be done with derivatives. Can someone prove it? Note that there are great chances for it to be true. So, this is the only step to a final solution to this superhard problem!

Till someone proves the last inequality, the solution is incomplete and even wrong. So, please, someone, prove this last inequality or disprove it!

[quote="harazi"]This monster would make jealous even Lagrangia:
   For any $ x\geq n-1$ we have the inequality
   $\frac{(1+x)(n+\frac{x^{n-1}-(n-1)^{n-1}}{(n-2)^{n-2}})-n}{n-1}\geq n+\frac{(x+1)^n-n^n}{(n-1)^{n-1}} $.[/quote]

 :lol:  :lol:

If this is true.. then this inequality is very nice!  :D

I just love this inequalities! :)

Well, I think it is true, and I think I have proved it. Anyway, the proof is quite difficult and uses lots of formulas, so I let you prove it!  :D

All this having been said, I think that we can state a theorem, which I don't know if it's known.
      For any $ n\geq k\geq 1$ we have $ inf\frac{a_1^k+a_2^k+...+a_n^k-n}{(a_1+a_2+...+a_n)^k-n^k}=\frac{1}{(n-1)^{k-1}} $ where the infimum is considered after all sistems $ (a_1,...,a_n)<>(1,1,1,...,1) $ with product 1. ( Popa Calin and Harazi).
    Just an observation: this is an even larger generalization of Turkevici inequality!

This is a NEW and BEAUTIFUL theorem.

Congratulations for your efforts to prove it.


Bebinam kasi midoone cheraa too bakhshe "Resources" masalehaaye kheyli az keshvarhaaro gozaashtand; vali masale haaye olampiaad iran ro nazashtand?

fekr konam hadeaghal marhale dovom haaye iran ro mishe oonjaa gozaasht na?

barobach; motshakeram az esteghbaaletoon :clap:

[quote="amirhtlusa"]etefaghan man daram sol haye 98/99 iran o latex mikonm mifresam be Mod. :)

hopefully soon[/quote]
bebin lotfan yaadet baashe yek tabaghe bandie dorosto hesaabi daashte baashe (marhal dovom; marhale sevom; TST1; TST2; ....)
daset dard nakone. mamnoon ;)

rastesh, man motmaen nistam. in soal a male koodoom marhaleh hast, in a mal e 98 e. age kasi midooneh begin, man 16 ta soal daram in 7 taye aveleshe :
1.Let $KL$ and $KN$ be tangent lines to the circle $C$ with $L,N\in C$.  choose $M$ on the extension of  $KN$ past $N$, and let $P$ be the second intersection of $C$ with the circumcircle of $KLM$.  Let $Q$ be the foot of the perpendecualar from $N$ to $ML$.  Prove that $\angle MPQ=2\angle KML$

2.Suppose an $n\times n$ table is filled with the numbers $0, 1,-1$ in such a way that every row and every column contains exactly one 1 and one -1. Prove that the rows and columns can be reordered so that in the resulting table each number has been replaced by its negative.

3.let $x_1,x_2,x_3,x_4$ be positive real such that:
$x_1.x_2.x_3.x_4=1$
Prove that:
$\sum_{i=1}^{4}x_i^3\geq max\{\sum_{i=1}^{4}x_i,\sum_{i=1}^{4}\frac{1}{x_i}\}$

4.let $ABC$ be an acute triangle and let $D$ be the foot of the altitude from $A$. Let the angle bisectors of $B$ and $C$ meet $AC$ at $E$ and $F$, respectively. If $BE=CF$, prove that $\triangle ABC$ is an isosceles triangle.

5.suppose $a,b$ are natural numbers such that:
$p=\frac{b}{4}\sqrt{\frac{2a-b}{2+b}}$ is prime. What the maximum possible value of $p$?

6. let $x,a,b$ be positive integers such that $x^{a+b}=a^bb$ prove that $a=x, b=x^x$.

7.Let $ABC$ be an acute triange and $D,E,F$ the feet of its altitudes from $A,B,C$ respectively. The line through $D$ parallel to $EF$ meets line $AC$ and line $AB$ and $Q$ and $R$ respectively.  Let $P$ the intersection of line $BC$ and line $EF$. Prove that circumcircle of $PQR$ passes through the midpoint of $BC$

darzemn age baro bach, harki soal ha ro male salyeh ghabl o dareh inja post konan. :)

soalhayeh olampiadhayeh 2 sal akhir keh ghablan dar bakhsh soalat bood,(bacheha post kardeh boodan)alan ham
fekr konam basheh faghat bayad yeh search kocholoo  kard!

[quote="amirhtlusa"]rastesh, man motmaen nistam. in soal a male koodoom marhaleh hast, in a mal e 98 e. age kasi midooneh begin, man 16 ta soal daram in 7 taye aveleshe :[/quote]

marhale dovom 1376:
soaale 1 rooze aval :
 http://www.mathlinks.ro/Forum/post-243401.html&highlight=4y^2+y#

soaale 2 rooze aval : hamoon 1 khodet.

soaale 3 rooze aval : hamoon 2 khodet.
.......................................

soaale 1 rooze dovom : hamoon 3 khodet

soaale 2 rooze dovom : hamoon 4 khodet

soaale 3 rooze dovom : hamoon 5 khodet

marhaleye 3 chanta soal dare?
tst1? tst2?


Show that $ 3^n\plus{}2\cdot 17^n$ is never a perfect square with $ n\in\mathbb{N}$.

Modulo $ 4$, the given expression is $ (\minus{}1)^n\plus{}2$. But a square is $ 0$ or $ 1$ modulo $ 4$, then $ n$ is odd.
It follows that $ n\equal{}2k\plus{}1$ for some non-negative integer $ k$ and the given expression is $ 3 \cdot 9^k \plus{} 2\cdot 17 \cdot 17^{2k} \equal{} 1\plus{}2 \equal{} 3 \mod [8]$.
But a square is $ 0,1$ or $ 4$ modulo $ 8$, hence result.

Pierre.

in modulo $ 8$
$ 3\cdot 9^k\plus{}2\cdot 17\cdot 17^{2k}\equiv 3\plus{}2\equiv 5 (mod\ 8)$ :maybe:
but really is the same  :rotfl:

[quote="pelao_malo"]Show that $ 3^n \plus{} 2\cdot 17^n$ is never a perfect square with $ n\in\mathbb{N}$.[/quote]
Suppose $ 3^n\plus{}2.17^n\equal{}x^2$
Consider mod 4 we have $ n\equiv 1(\mod 2)$
So $ n\equal{}2k\plus{}1$
So $ 17|x^2\minus{}3.(3^k)^2$
But $ (\frac{3}{17})\equal{}(\frac {17}{3})\equal{}(\frac{2}{3})\equal{}\minus{}1$
It show the contradition .
It mean that $ 3^n\plus{}2.17^n$ can not be a perfect square.


modular arithmetic
number theory unsolved
number theory

Prove that, for any integer $ n$, the number $ a\equal{}5^{5^{n\plus{}1}}\plus{}5^{5^n}\plus{}1$ can not be prime.

$ a\equal{}k^5\plus{}k\plus{}1$ where k=5^(5^n) :wink:

[quote="moldovan"]Prove that, for any integer $ n$, the number $ a \equal{} 5^{5^{n \plus{} 1}} \plus{} 5^{5^n} \plus{} 1$ can not be prime.[/quote]

$ n$ even $ \implies$ $ 5^n \equal{} 1\pmod 3$ and $ 5^{5^n} \equal{} 5\pmod{31}$ and $ 5^{n \plus{} 1} \equal{} 2\pmod 3$ and $ 5^{5^{n \plus{} 1}} \equal{} 25\pmod{31}$ and so $ a \equal{} 0\pmod{31}$

$ n$ odd $ \implies$ $ 5^n \equal{} 2\pmod 3$ and $ 5^{5^n} \equal{} 25\pmod{31}$ and $ 5^{n \plus{} 1} \equal{} 1\pmod 3$ and $ 5^{5^{n \plus{} 1}} \equal{} 5\pmod{31}$ and so $ a \equal{} 0\pmod{31}$

So $ a\equal{}0\pmod {31}$ and, since $ a>31$ $ \forall n\geq 0$, $ a$ is not prime.

Posted another time also with pco  :wink:
If $ a_0,a_1,a_2,\ldots,a_{p \minus{} 1}$ is a complete residue in $ \mathbb{F}_p$ for some $ p \in \mathbb{P}$ then $ \displaystyle \sum_{0 \le i \le p \minus{} 1}{x^i} \mid \sum_{0 \le i \le p \minus{} 1}{x^{a_i}}$ (in fact 31=1+5+25)


Which triplet of numbers has a sum NOT equal to 1? \\ \\
$ \text{(A)}\ \frac{1}{2}, \frac{1}{6}, \frac{1}{3} \qquad \text{(B)}\ 2,\minus{}2,1 \qquad \text{(C)}\ 0.1,0.3,0.6 \qquad \\ \text{(D)}\ 1.1,\minus{}2.1,1.0 \qquad \text{(E)}\ \minus{}\frac{3}{2},\minus{}\frac{5}{2},5$ \\ \\
Select the most appropriate letter.

You can immediately see that $ 1.1\plus{}(\minus{}2.1)\plus{}1.0\equal{}0$. This makes the answer $ (\text{D})$

[quote="quattayangel"][quote="GameBot"]Which triplet of numbers has a sum NOT equal to 1? \\ \\
$ \text{(A)}\ \frac {1}{2}, \frac {1}{6}, \frac {1}{3} \qquad \text{(B)}\ 2, \minus{} 2,1 \qquad \text{(C)}\ 0.1,0.3,0.6 \qquad \\
\text{(D)}\ 1.1, \minus{} 2.1,1.0 \qquad \text{(E)}\ \minus{} \frac {3}{2}, \minus{} \frac {5}{2},5$ \\ \\
Select the most appropriate letter.[/quote]

The answer can also be C because 0.1+0.3+0.6=0.9 and 0.9 is not equal 1[/quote]
0.1+0.3+0.6=1.0


[quote="FelixD"]$ \sum a^{1996} \le \sum a$, as $ 0 \le a \le 1$. Let $ \sum a \equal{} x$. We have to prove $ (x \plus{} 1)^2 \ge 4x \Leftrightarrow (x \minus{} 1)^2 \ge 0$.[/quote]

Cute solution.
Thanks.


quadratics
abstract algebra
number theory proposed
number theory

Solve the equations in $N$ :
 a) $3^x =2y^2+1$
 b) $2^x =3y^2+1$

surely you do not mean both hold at the same time?

The second one: for $x>2$ 8 divides both sides, but
mod 8 the right hand side can be congouent only to $3\times 0+1\equiv 1$, $3\times 4+1\equiv 5$ or $3\times 1+1\equiv 4$.
For x=2 we have y=1 and for x=1 there's no solution.

Nice solution , Kondr !
 For the first one , I think $x=y=1$ or $x=y=2$  ;)  but I haven't a solution

Unfortunately x=5 and y=11 is also a solution. There are no other solutions for x<200 (by Maple). I have no idea how to solve this. The method i used above would need congguences mod 729 and when testing all quadratic residues for this module, i found some numbers for which the congrouence
$0\equiv 2y^2+1 (mod 729)$ holds.


vector
geometry
induction
linear algebra
linear algebra unsolved

This "exercise" is in my mind really a fun problem, somewhat confusing since it seems so obvious. I heard it as a puzzle some years before I found it in Artin's book. There's more than one way to solve this.

[b]Prove that a finite-dimensional vector space over an infinite field is not the union of finitely many proper subspaces.[/b]

"Proper subspace" just means "contained in but not equal to the vector space". Note that "union" means just that - not "linear span", but union as in set theory.

Forgive me if this is a bit high-powered, but as a variety a given finite-dimensional vector space of dimension $ d$ is a variety of dimension $ d$ (in the sense of algebraic geometry) whereas the union of finitely many of its proper subspaces has dimension at most $ d\minus{}1$ (again in the sense of algebraic geometry).

"The union has dimension at most $ d\minus{}1$" is probably dependent on the field being infinite, right? What exactly is the argument you're using here? Also, I wonder if the dimension argument holds in characteristic $ p$. The statement is correct over any infinite field, including ones of finite characteristic.

And yes, it's a bit high powered  :)

The algebraic geometry notion of dimension is a very tricky beast.

A low-powered method: induction on dimension. Suppose $ S$ is a finite set of subspaces, and its union is the whole vector space $ V$. We want to show that some element of $ S$ is itself $ V$.
In dimension $ 1$, this is obvious.
In dimension $ n$ for $ n>1$, there are infinitely many subspaces of dimension $ n\minus{}1$. Choose one such subspace $ U$ which is not in $ S$. The intersections of the elements of $ S$ with $ U$ are subspaces of $ U$; by the inductive hypothesis, one is itself $ U$. Since $ U$ is not an element of $ S$, it must be properly contained in an element. That element is the whole space $ V$, and we are done.

[quote="jmerry"]In dimension $ n$ for $ n > 1$, there are infinitely many subspaces of dimension $ n \minus{} 1$. [/quote]
This is the one and only place where jmerry used the hypothesis that the field is infinite.

Of course, the conclusion of the original problem is false over a finite field.

Another method: $ V$ is isomorphic to $ F^n$. Consider vectors of the form $ (1,t, t^2, \ldots, t^{n\minus{}1})$ for $ t \in F$. Since the field is infinite, there are infinitely many of them, however every proper subspace can only contain $ n\minus{}1$ of them because every set of $ n$ of these vectors is independent (Vandermonde). That's it!

This method is less elegant than jmerry's since it relies on identifying $ V$ with $ F^n$, but it has the advantage that it can be used to prove an analogous quantitative statement over finite fields.

The problem reminds me of the following result:

[b] If $ U, W$ are two vector subspaces of $ V$ then $ U \cup W$ is a vector subspace if and only if $ U \subseteq W$ or $ W \subseteq U$. [/b]

I think we could use similar argument to prove the generalized one: (don't know if it is true but I cannot come up with any counterexample)

[b] If $ U_1, U_2, \dots, U_n$ are $ n$ vector subspaces of $ V$ then $ \bigcup_{i\equal{}1}^{n} {U_i}$ is a vector subspace if and only if one of the $ U_i$ contains all the others. [/b]

With that generalization, the original problem of [b] randomgraph [/b] follows.

[quote="mathangel"]The problem reminds me of the following result:

I think we could use similar argument to prove the generalized one: (don't know if it is true but I cannot come up with any counterexample)

[b] If $ U_1, U_2, \dots, U_n$ are $ n$ vector subspaces of $ V$ then $ \bigcup_{i \equal{} 1}^{n} {U_i}$ is a vector subspace if and only if one of the $ U_i$ contains all the others. [/b]

With that generalization, the original problem of [b] randomgraph [/b] follows.[/quote]

Counterexample: Let $ V$ be a finite-dimensional vector space over a finite field and let $ U_i$ be all proper subspaces (or all 1-dimensional subspaces, or all $ n\minus{}1$-dimensional subspaces, or a mixture of some of those..)


We say about xi (i from 1 to n) that they are Q-conjugates if they are the roots (in C) of a polynomial in Q[x]. Let f be a rational polynomial of degree n with roots xi (i from 1 to n) and let ri (i=1->n) be rational numbers s.t. xi+ri are Q-conjugates. If f is irreducible in Q[x], show that r1=r2=..=rn.

[quote="grobber"]We say about xi (i from 1 to n) that they are Q-conjugates if they are the roots (in C) of a polynomial in Q[x]. [/quote]

Must that polynomial have degree n or less?

I don't really know :). I took it to have degree n. That's how I understood it at first, and that's how I think it is. It seems natural..


bacheha inja chera inghadr bi mohtava shode?

say konid be jaye bahsaye hashiyei bishtar soal bezarid va hal konid! :wink:

baraye shoru sai konid in soal ro hal konid:
$\sum{x_{i}}^{2}=1 \ (mod p)$ in moadele chand javab darad.($p$ aval va fard)

zigma $i=1$ ta $n$ hastesh.
in soal ba jame jacoobi hal mishe.
khob man ye soal midam:
ye jadvale n*n darim.
do nafar shoru mikonan dar in jadval domino gozashtan be tori ke ham pooshani nadashte bashand hala age kasi natoone edame bede bakhte hala ki estrateji bord dare?
in soal baraye n haye zoj ke badihiye baraye n haye fard javab ra biyabid.

salam.kasi mitooni raaje be darsaaye doreye emsaal tozih bede?
ya manaabe e mofid ro moarrefi kone?
baa tashakkor

man baraye in soal ye rahe hale sakht daram agar bekhain mitunam post konam

Manam moafegham inja kheili , kheili bad shode . hala manam 2ta soal mizaram ke soale khoobian va baraye alaghemandane nazarie adade o jabre
sabet konid moadeleye zir dar adade sahih binahayat javab darad :
1)$x^{3}+y^{3}+z^{3}+t^{3}=1999$
2)agar $a+b+c=1$ sabet konid
$10(a^{3}+b^{3}+c^{3})-9(a^{5}+b^{5}+c^{5})\geq1$

laaghal soalaro doone doone befrstid,yeni sab konid javabe ye soal dade beshe bad soale bado bezarid...
(baraye soale namosavie:[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=362838#362838[/url])

sabet konid moadele sayale $x^{2}+y^{2}+z^{2}+u^{2}+v^{2}=xyzuv-65$ javabi darad ke dar an  tamame adaade $x,y,z,u,v$ az $1988$ bishtar hastand..
nemidonam sakhte ya asoonehaa yani fek nakardam :wink:

Baraye soale sepehr : avval inke $(1,2,3,4,5)$ va $(2,3,4,5,119)$ do javab baraye in moadele hastand . hala age $(x,y,z,u,v)$ tooye in moadele sedgh konan az oonja ke $x^{2}-xyzuv+y^{2}+z^{2}+v^{2}+u^{2}+65=0$ ye moadeleye darajeye dovvom bar hasbe $x$ hast va majmooe rishe ha $yzuv$ hast pas $yzuv-x$ ham risheye digeye in moadelast va maloome ke az hameye $x,y,z,u,v$ ham bozorgtare . hala mitoonin ba inkar ye donbale javab besazin ke tooye har marhale koochiktarineshoono az ooni ke too marhale ghabl bozorgtar boode bozorgtar konin . masalan az $(1,2,3,4,5)$ be $(2,3,4,5,119)$ va az javabe jadid be $(3,4,5,119,7138)$ beresin . pas bad az chandta marhale be ye javabi ke masale khaste miresim .
roo soale man kasi ideayi nadare ??
ye rahnamayi mikonam oonam ine ke farz konin $t=-1$ va sabet konin moadeleye $x^{3}+y^{3}+z^{3}=2000$ binahayat javab dare :)


Seien $ a$ und $ b$ reelle Zahlen. Die Folge definiert:
$ a_0 \equal{} a$; $ a_1 \equal{} b$; $ a_n \equal{} \frac {1}{2}(a_(n \minus{} 1) \plus{} a_(n \minus{} 2))$ fur $ n\geq 2$.

Man beweise, dass die Folge konvergiert und bestimme ihren Grenzwert.

Fitim Dika

Hi Fitim.

Versuche es mit em Ansatz
$ b_n\equal{}a_{n\plus{}1}\minus{}a_n$

Dann kannst du das ganze nach ein bisschen Rechenarbeit auf eine geometrische Summe zurückführen.

EtotheX


How many positive multiples of 5 that are less than 100 have a units digit of 5?

This is simply the number of odd integers from $ 1$ to $ 19$. Since the $ x$th positive odd number is $ 2x\minus{}1$, we have $ 2x\minus{}1\equal{}19 \implies x\equal{}\boxed{10}$.


Let $ A\in N$ be given. Permuting the figures of $ A$, we get $ B$. Suppose that $ A \minus{} B \equal{} 111...1$ which has $ n$ figures. Find the least value of $ n$.

$ A\minus{}B \equiv 0$ mod 9

Hence, we must have $ n$ is a multiple of $ 9$

$ 987654320\minus{}876543209\equal{}111,111,111$

Hence the smallest value is $ n\equal{}9$


[color=darkblue]Prin amabilitatea tanarului meu prieten, profesorul [b]Adrian Troie[/b] din M.Ed.C.,
iata [b][u]Calendarul concursurilor de Matematica din Romania[/u][/b].
Rog Administratorul acestui Site sa plaseze in amonte acest topic.
Va doresc mult succes ![/color]

[color=darkblue]Multumesc., mi-a facut o deosebita placere.
Voi sunteti speranta noastra in viitorul acestei tari.[/color]

Stie cineva daca se mai tine concursul [b]"gheorghe vranceanu"[/b], bacau? Ar fi trebuit sa aiba loc in decembrie, dar s-a zis ca din cauza lipsei fondurilor se va amana pana in februarie...si nu a avut loc.
Ar fi fost o buna pregatire pentru nationala.

concursul de la bacau a avut loc, a fost chiar si cu proba de baraj, subiectele de la aceasta proba fiind puse de maky aici : [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=120264[/url]

cosmin, ma refer la "vranceanu", nu la "vranceanu procopiu". Primul e de mate, al doilea de mate-fizica, care pe langa subiectele propriu-zise are si subiecte de baraje (iti alegi tu obiectul). Subiectele postate de maky sunt de la baraj-mate-liceu, care a ramas singurul baraj la mate, pt ca incepand din 2006 se tine doar pt liceu.

[quote="mychrom"]Multumim mult pentru efortul dumneavoastra, domnule Nicula.[/quote]

Ma alatur si eu la multumiri si as adauga la lista "Concursul interjudeţean de matematică Traian Lalescu", programat pe 23-25 martie 2007, la care participa elevi din judetele Arad, Caras-Severin, Hunedoara si Timis (in principiu, primii 5 clasati la etapa judeteana a Olimpiadei de Matematica, la fiecare clasa V-XII). Mod de desfasurare - individual. Anul acesta va fi editia XXI.

Daca cineva face rost de lista pentru anul acesta.. ar putea sa o puna aici?

Multumiri!

Cand are loc anul acesta concursul "MODUS VIVENDI" din R. Valcea ?!
Tot pe 24 februarie sau mai devreme ?!

Scrie pe site-ul lor : http://mircea.unet.ro

Numai bine!

Bonjour !  :)

Sunt [b]Julian Muscalu[/b], [b]prof. Mate-Info[/b], participant la etapele finale ale [b]ONM[/b] (anii [b]1984-1986,[/b] în 1987 nu m-am calificat dintr-un motiv "mai special"), membru al [b]Lotului Olimpic lărgit[/b] ([b]1985[/b] - pt. OIM Helsinki). La [b]Profil[/b] se găsesc mai multe detalii (în [b]G.M.[/b] am semnat la vremea respectivă [b]Muscalu Cătălin[/b], folosind celălalt prenume).

Fratele meu [b]Adrian Muscalu[/b] a fost și el participant la fazele finale ale ONM și prezent în [b]Lotul Olimpic lărgit în 1978[/b] (OIM România). El era în generație cu [b]Nistor, Zaharescu, Cavachi[/b] ...etc ... (detalii la [b]micul site web[/b] pe care l-am pus la Profil).

[b]Caut diferite modalități de a-mi procura diverse cărți[/b] (pe care nu le găsesc pe Internet - la Anticariatele găsite nu mai sunt disponibile cărțile respective și nu vad ediții noi pentru cărțile respective.
Un exemplu: ediția coordonata de Ioan Tomescu a olimpiadelor naționale 1950-1990 ... și mai sunt și altele).

Pe Internet am văzut ca (ar) exista o ediție [b]"Romanian Mathematical Competitions" (RMC) din 1996 - 2008[/b]. Dar nu prea găsesc detalii pe nicăieri despre cum as putea cumpăra această carte (nu găsesc o librărie on-line ..etc...)

Am văzut că [b]anul acesta ONM-ul este la Iași.[/b]
[b]Am și eu niște probleme de propus[/b] (pt [b]Baraje sau clasa a 10-a, în principiu[/b]). Evident nu pot da detalii aici, o sa contactez câteva persoane care participă la ONM si care pot propune juriului respectivele probleme.

In ultimii 20 de ani m-am ocupat mai mult de programare (C++, [b]Java[/b], mai recent JavaScript și PHP/MySQL  ... am studiat (și aplicat) în anumite perioade Delphi și Fox, care acum vad ca au cam dispărut și sunt înlocuite de Oracle - o sa studiez și Oracle Agil, am înțeles ca asta-i fain... enfin...

Dar [b]mereu am simțit nevoia sa revin la Matematica de performanță[/b] (cea de la clasă, aici unde predau în acest moment, nu ma satisface deloc - de 10 ani predau la o școală generală, în Cartojani-Giurgiu, eu locuind în Gratia-Teleorman, acasă, la 2 km de școală.
Am mai predat 2 ani la o alta scoală generală pe aici (tot Mate), plus Info la 3 licee, la clase de Informatică, în București: Jean Monet, apoi Spiru Haret și apoi Tudor Vianu).
In Mate [b]zona universitara[/b] ma interesează [b]"Combinatorica și Teoria Grafurilor"  [/b]

Iar acum resimt aceasta dorință mai mult decât oricând, chiar dacă am 40 de ani. Ma interesează (măcar) problemele care se dau și se discuta, dar și ce comportament au elevii de azi în fața acestor probleme...  :)

Nefăcând meditații (fiind mereu ocupat cu "studiul" mai multor lucruri) a trebuit să părăsesc Bucureștiul.
Noroc cu Internetul.
Anul trecut am cumpărat câteva cărți la librăriile on-line.
Sper sa reușesc și acum sa găsesc ce caut.

In afara de [b]AMC[/b] și cartea coordonată de I. Tomescu mai caut (și nu găsesc) și "Data Structures" (Tomescu) dar și alte cărți legate de [b]Combinatorica și Teoria Grafurilor[/b] și de [b]Olimpiadele și concursurile interjudețene, naționale și internaționale[/b].

Eu am mai multe PDF-uri (majoritatea în engleza, și nu e problemă la mine cu engleza sau franceza) dar e MULT mai convenabil sa lucrez după o carte decât după niște foi scoase la imprimanta...

Poate cineva cunoaște niște [b]LINK[/b]-uri unde as putea comanda aceste cărți (sau cărți din aceste domenii).

OK. Mersi mult !

Bye !

Julian _____


Let $a_1,...,a_n$ be given real numbers which are not all zero. Determine the least value of $x_1^2+...+x_n^2$ where $x_1,...,x_n$ are real numbers satisfying $a_1x_1+...+a_nx_n=1$

Cauchy: \[\sum x_i^2 \sum a_i^2 \geq \left(\sum x_ia_i\right)^2.\]

Since $\sum x_ia_i = 1$ and $\sum a_i^2$ is fixed, that gives us the minimum.


Find a set of 10 different positive integers such that each number is a divisor of the sum of the integers.

[hide]By a bit of trial and error, {1,2,3} has the property that every number is a factor of the sum 6.  Then if we add 6 to the set, the total sum will be doubled, so 6 will be a factor of the sum and the rest will remain factors as well.  We can then add 12, then 24, and so on, since each time the new sum will be twice the old sum (and thus will remain a multiple of each old number in the set) and will also be a multiple of the newly added number.

Thus we get one possible set as {1, 2, 3, 6, 12, 24, 48, 96, 192, 384}.[/hide]


If $ 935712 \times N$ is a perfect cube for some positive integer $ N$, find the minimum value of $ N$.


Stim ca orice dreptunghi e inscriptibil intr-un cerc. Este raza cercului circumscris acestuia egala cu jumatate din diagonala lui  :huh: ? Multumesc.

dreptunghiul e format din 2 triunghiuri dreptunghice care au latura comuna ipotenuza si anume una din diagonalele
dreptunghiului(depinde pe care consideri).
se stie ca masura unghiului ce subintinde un arc este jumatete din masura arcului.
cum masura arcului a carei coarda este ipotenuza(diametru) e 180 => ca unghiul coresp va fi 90.


[hide = why did I waste my time on this?]Summing we see that:
$$(x_1x_2 - x_1x_4)^2 + (x_2x_3 - x_2x_5)^2 + ... + (x_5x_2 - x_5x_4)^2 \leq 0$$
from which we have that:
$$x_1 = x_2 = x_3 = x_4 = x_5$$[/hide]

Sum and expand to get to a sum of "squares (made of difference of products of $x_i$)" $\leq 0.$ So every "square" is a zero and every $x_i$ is equal.


inequalities
combinatorics proposed
combinatorics

Let $ S\equal{}\{a_1,a_2,\dots,a_n\}$,where $ a_i\in\mathbb{N}$,for $ i\equal{}1,2\dots,n$.We will call set $ S$ -"free" if all subsets of $ S$ have diffrent sums.If $ S$ is "free" then find the minimum value of $ a_1^2\plus{}a_2^2\plus{}\dots\plus{}a_n^2$?

[quote="yunxiu"]A={1,2,4,...,2^(n-1)} is free.
For any free B={a1,a2,...,ak},MaxB≥2^k-1.
If a1≤a2≤...≤an, {a1,a2,...,ak}is free for 1≤k≤n,
So ∑a^2≥1+2^2+4^2+...+2^(2n-2)=1/3[4^n-1].[/quote]
Your solution is incorrect,because,as you said,if $ B\equal{}\{a_1,a_2,\dots,a_k\}$ is free,then $ max\{B\}\geq 2^{k\minus{}1}$,which is wrong,because
the set $ \{3,5,6,7\}$ is free,but the max element is less than $ 2^{3}$

As i said before,this problem was at Kazakhstan IMO TST. Though this problem is not so tough, it took me two hours to find the proof
Answer:$ \frac {2^{2n} \minus{} 1}{3}$.And the equality occurs if $ a_i \equal{} 2^{i \minus{} 1}$.
Solution:
Let $ A \equal{} a_1 \plus{} a_2 \plus{} \dots \plus{} a_n$ and $ Q \equal{} a_1^2 \plus{} a_2^2 \plus{} \dots \plus{} a_n^2$.Also let $ 0 \equal{} d_0 < d_1 < d_2 < \dots < d_{2^{n} \minus{} 1} \equal{} A$,where $ d_i$ are elements sums in all possible subsets of $ S$
Here we notice that $ \boxed{A\geq 2^{n} \minus{} 1}$.
Consider the sum $ P \equal{} d_0^2 \plus{} d_1^2 \plus{} \dots \plus{} d_{2^{n} \minus{} 2}^2 \plus{} d_{2^{n} \minus{} 1}^2$.
Lemma:
$ P \equal{} 2^{n \minus{} 2}(A^2 \plus{} Q)$.
Proof is obvious.If someone needs a proof,i can post it.
Now we will use the following inequality:
If  $ \frac {S}{2}\geq d\geq d'$,then
$ d^2 \plus{} (S \minus{} d)^2\leq (S \minus{} d')^2 \plus{} d'^2$.
Now we will consider two cases:
1.$ n \equal{} 2k \plus{} 1$:
We know that $ d_i \plus{} d_{2^{n} \minus{} 1 \minus{} i} \equal{} A$,hence
if we change $ d_{2^{n \minus{} 1} \minus{} 1}$ to $ k$ then authomatically $ d_{2^{n \minus{} 1}} \equal{} k \plus{} 1$ and $ P$ will become less.
In similar way,we change $ d_{2^{n \minus{} 1} \minus{} 2}$ to $ k \minus{} 1$ and $ d_{2^{n \minus{} 1} \plus{} 1} \equal{} k \plus{} 2$,
$ d_{2^{n \minus{} 1} \minus{} 3}$ to $ k \minus{} 2$ and $ d_{2^{n \minus{} 1} \plus{} 2} \equal{} k \plus{} 3$ and so on...
So

$ P\geq (k^2 \plus{} (k \plus{} 1)^2) \plus{} ((k \minus{} 1)^2 \plus{} (k \plus{} 2)^2) \plus{} \dots \plus{} ((k \minus{} 2^{n \minus{} 1} \plus{} 2)^2 \plus{} (k \plus{} 2^{n \minus{} 1} \minus{} 1)^2) \plus{} (2k \plus{} 1)^2$
$ \equal{} 2k^2 \plus{} 2(2^{n \minus{} 1} \minus{} 1)k \plus{} (2^{n \minus{} 1} \minus{} 1)^2 \plus{} (4k^2 \plus{} 4k \plus{} 1) \plus{} 2(\sum_{i \equal{} 1}^{2^{n \minus{} 1} \plus{} 2}k^2 \plus{} i^2)$
$ \equal{} 1 \plus{} (2^n \plus{} 2)k^2 \plus{} k(2^n \plus{} 2) \plus{} (2^{n \minus{} 1} \minus{} 1)^2 \plus{} \frac {(2^{n \minus{} 1} \minus{} 2)(2^{n \minus{} 1} \minus{} 1)(2^{n} \minus{} 3)}{3}$
As we have noticed in lemma $ P \equal{} 2^{n \minus{} 2}(A^2 \plus{} Q)$,hence
$ P \equal{} 2^{n}k^2 \plus{} 2^{n}k \plus{} 2^{n \minus{} 2} \plus{} 2^{n \minus{} 2}Q$,thus
$ 2^{n \minus{} 2}Q\geq 1 \plus{} 2k^2 \plus{} 2k \plus{} (2^{n \minus{} 1} \minus{} 1)^2 \minus{} 2^{n \minus{} 2} \plus{} \frac {(2^{n \minus{} 1} \minus{} 2)(2^{n \minus{} 1} \minus{} 1)(2^{n} \minus{} 3)}{3}$
Now using that $ 2k \plus{} 1\geq 2^{n} \minus{} 1\Rightarrow k\geq 2^{n \minus{} 1} \minus{} 1$,thus
$ 2^{n \minus{} 2}Q\geq\frac {2^{3n \minus{} 2} \minus{} 2^{n \minus{} 2}}{3}$,so
$ Q\geq\frac {2^{2n} \minus{} 1}{3}$.
2.$ n \equal{} 2k$.
I will not write details for this case,but if we use the same inequality,we will obtain
$ Q\geq\frac {2^{2n} \plus{} 3\cdot2^{n} \plus{} 2}{3}\geq\frac {2^{2n} \minus{} 1}{3}$,so the minimum for this case is greater than for the first and we are done. :)


[quote="charikaar"]hello,

does anyone know how to solve this question.

Rob reduced phone price by 30%. He sold 60% more phones in sale.

calculate the percentage increase in its takings.

Thanks for your help.

Charikaar[/quote]

Umm...I really have no idea. This is just a guess, but I don't think it's right.

[hide]
I'm going to pretend that he had 100 phones selling for 100 dollars each. (I don't think you can just assume like that in this problem, but I don't know what else to do.) So reducing the price by 30% means that they are now 70 dollars each. If he was going to sell 100, then an increase by 60% would mean 160 phones sold.

Originally, would've sold 100 phones for 100 dollars each. That's 10,000 dollars.

Now, he's selling 160 phones for 70 dollars each. That's 11,200 dollars.

Percent of increase = change/original
1200/10000=x/100
x=12

Is 12% right? I dobut it...I think I did way too much assuming.
[/hide]

[quote="charikaar"]hello,
does anyone know how to solve this question.
Rob reduced phone price by 30%. He sold 60% more phones in sale.
calculate the percentage increase in its takings.

Thanks for your help.

Charikaar[/quote]

I believe it depends also in gain rate. If he reduces 30%, probably is because his ganing rate is at least 35% or so.

If not, then, you may try this. His takins are $t=np$ where $t$ are takings, $n$ is the number of phones sold, and $p$ is the price of each individual phone. Then you may say that $n$ changes to $1.6n$ (an increment of the 60% in the number of phones) and $p$ changes to $0.7p$ (because there was a reduction of 30%) therefore $t$ changes to $0.7p\cdot1.6n=1.12np=1.12t$, therefore, you have an increase of 12% in his takings.

Best,

[quote="shinwoo"]oooooooo....

wouldn't that depend on the day? luckyday for selling 60% more or bad day for selling nothing...

I don't think I can solve this problem.. I think it should go up to much higher leveled forum....

sorry... :blush:  :([/quote]

this was one of my exam question and I got 19% and i think i am wrong because i did it exactly the same way as above but subtracted the number incorrectly(11200-10000=1200 meant to and i had something else lost 5 easy points out of 15)

I think 12% is the correct answer.

Thanks for your help.

Charikaar

i mean i attempted the question the same way as SoccerBrainy40

charikaar

[hide="solution"]say n is the original price, m is the sale percentage.

since he gained 60% sale after reducing the price 30%,

it would be $\left(1 - .3 \right)n \left(1 + .6 \right)m$
$ = 0.7n * 1.6m$
$ = 1.12mn$

which is .12 raised from 1mn, meaning 12%
[/hide]

oops too late.


Sign up here for a mock team round for Mathcounts practice. If you want to join, PM me and have the PM include your score at chapters/states/nationals last year, depending how far you got. Or, you can use other info, like AMC scores. I'll try to put together even teams. Thanks!

I'd like to participate. I've never been in any official contest. You can put me in the group of getting started people

lol.. sounds fun.. ahh.. the good ole MathCounts days...

AMC 10A: 118
AMC 10B: 125.5

.. i think.. (im very forgetful.. lol)

I think we should just do this by the scores of the Mock Mathcounts contest. If you didn't take it, I will try to figure out your level by your  posts on the forums and place you in a team. Thanks!

When will it be? I would like to sign up, but I cannot positively gurantee that I will be on until I know all the info about it. Thanks

I'll do it. But could you email me a few days beforehand? I don't always check this place often.

How will people discuss the questions?

Do you realize that the mock happened 16 years ago?

[quote=RedFireTruck]BUMP!!!!![/quote]

[quote=Eric2009]can i do it? this is my first time joining a mathcounts team (mock or not)[/quote]

You two are either trolling or being bad citizens of aops. Reported.

I think Eric2009 didn't do anything wrong - it was just an honest mistake and they didn't see the timestamps! :maybe:

[quote=Iamnobody]@above, if that is true, then his record is still clear. He will probably just recieve a warning message.[/quote]

True! (But RedFireTruck is definitely a troll!!! :mad:)

[quote=falcon22][quote=franzliszt]RedFireTruck actually makes some legit posts in the Algebra Marathon. No idea why (s)he is bumping like mad.[/quote]

Really??!!! Well, that seems impossible with the kind of posts they are making now!!![/quote]

I mean
u can go see for urself
it's a public thread


Find the $n$ th term of the sequence $\{a_{n}\}$ such that $a_{1}=a\ (\neq-1),\ a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{1}{a_{n}}\right)\ (n\geq 1).$


geometry
geometric transformation
reflection

I'm a senior undergraduate math major who is looking to attend graduate school for operations research.  I've gone through the application process and the grad programs to which I applied have started making their decisions.  Because O.R. is rather specialized as far as math goes, I haven't been able to find helpful literature that recognizes high quality O.R. programs.  What do you make of the programs below?  And do you think that the quality of a grad school's math program in general highly reflects the quality of an O.R. program?  Any help is appreciated.

UNC Chapel Hill
William and Mary
U of Delaware
Cornell

Thanks :!:

I know next to nothing about OR, but I've known multiple people to speak well of Cornell's OR program.


My state comp is on march 7th. I have a pretty good chance of making top ten and (maybe) nats. I also came down with some throat infection on Friday and it seems like its progressing into a fever. I really want to use my remaining time to review so I won't fail on the test unneccessariliy. Today was the first day where it seems like my sickness really weighed me down in school and tired me out. Should I review some math, or try to get better first?

Thanks

[b]Your health should always be your first priority.[/b] Anything else comes after that: don't worry about studying that much; the most important thing is that you [b]try your best[/b]. You should maybe take a few days off of school to make it easier to get better (and, unless it makes your sickness worse, you have lots of study time! :) ).

My opinion: (Not speaking from experience)
Practicing while tired will most likely stress you out.
When you are sick, you are tired.
Stress doesn't help you get better, and being sick during the competition will drag you down.
So get better. Though wait for some other people to post before making your decision.

Thanks guys, I'm just not sure if maybe I should just keep doing my normal life and somehow my immune system will take care of the virus by itself but if I calculate wrong, the consequences will be huge at state.

happened to me at chapter and i managed second only losing by a point....stay home...guys health is a priority but mc only comes once.......use it to ur advantage..i was gonna miss a week of school anyways for mathcounts state to prepare...but now u have an execuse......and studying while ur sick makes not much of a difference just give urself more breaks and less pressure if u get the questions wrong......

i may sound pathetic but then again, im pathetic


modular arithmetic
quadratics
number theory unsolved
number theory

Prove that for each prime $p \ge 7$, there exists a positive integer n and integers $x_1,\ldots,x_n,y_1,\ldots,y_n$ not divisible by p which satisfy
$\begin{array}{c}\.
x_1^2+y_1^2 \equiv x_2^2 \pmod p \
x_2^2+y_2^2 \equiv x_3^2 \pmod p \
\vdots \
x_{n-1}^2+y_{n-1}^2 \equiv x_n^2 \pmod p \
x_n^2+y_n^2 \equiv x_1^2 \pmod p \
\end{array}$

Lemma: for a prime $p\ge 7$ there is $a\in\mathbb Z_p\setminus\{0,1\}$ s.t. $a^2-1$ is a quadratic residue.

Proof: Take $g\in\mathbb Z_p\setminus\{0,1\}$ s.t. $g^2+1\ne 0$ (there is such a residue because $p\ge 7$). Then $a=\frac{g^2+1}{2g}$ satisfies the condition.

We now take $n=p-1$ and $x_i=a^{i-1},\ \forall i\in\overline{1,n}$.


function
induction
inequalities
limit
topology
real analysis
real analysis unsolved

I'm studying for an upcoming analysis test and the book I'm using, INtroduction to analysis by maxwell rosenlicht, is so crappy b/c it doesn't explain anything clearly and the problems are overly difficult and tedious w/o any hints or pointers or solution. I feel like I'm blindly trying to work problems and I don't even know if I'm on the right step or not. Can anyone helpe me with the following problems?

1. Show that the sequence of functions:

$\sqrt{x}, \sqrt{x+\sqrt{x}}, \sqrt{x+\sqrt{x+\sqrt{x}}}, \ldots$

on $\{x \in R: x \ge 0\}$ is convergent and find the limit function.

2. Show that the sequence of functions $x, x^{2}, x^{3}, \ldots$ converges uniformly on $[0,a]$ for any $a \in (0,1)$, but not on [0,1].

3. Show that the seqence of functions $x(1-x), x^{2}{(1-x)}, x^{3}{(1-x)}, \ldots$ converges uniformly on $[0,1]$.

4. Prove that the limit of a uniformly convergent subsequence of bounded functions (from one metric space to another) is bounded.

Thanks.

Hi,
Here is an idea:
1) Use the recurrence: $x_{n+1}=\sqrt{x+x_{n}}$, $x_{0}=x$, discussion after the values of x, the limit is $\frac{1+\sqrt{1+4x}}{2}$.
2. let $f_{n}=x^{n}$. On $[0,a]$ with $a<1$ the limit function is 0 and you have that $f_{n}(x)<a^{n}\rightarrow 0$, and you are done.
On the interval $[0,1]$ the limit function is $f(x)=0$ for $x\neq 0$ and $f(0)=1$ and observe that the sequence of functions is continuous on the interval $[0,1]$ while the limit function in not , therefore the convergence is not uniform.
3) Find the maximum of $x^{n}(1-x)$ on $[0,1]$, and simply observe that it is a sequence which converges to 0 , and you are done.
4) This is classic.
Didi

didilica, I know how to find the limit function for the first one, but can you elaborate on how to show that the $f_{n}= \sqrt{x+f_{n-1}}$ actually converges?

Hi,
Sorry for responding to you so late. Here is my approach to the first problem:

Definitely that if $x=0$ then $x_{n}=0$ so there is nothing to prove. Let $x\neq 0$. By induction show that the sequence is increasing. We have $x_{0}=\sqrt{x}<\sqrt{x+\sqrt{x}}=x_{1}$. Assume that $x_{n}<x_{n+1}$.  Adding x to both sides of the preceding inequality and taking the square roots you get that $x_{n+1}<x_{n+2}$, so you are done. Thus the sequence increases. Let $L=\frac{1+\sqrt{1+4x}}{2}$.
Next, by induction you prove that $x_{n}\leq L$. First observe that $x_{0}=\sqrt{x}<L$,      $2\sqrt{x}-1\leq \sqrt{1+4x}$
Next assume that $x_{n}\leq L$ and show that $x_{n+1}\leq L$. This follows imediately since
$x_{n+1}=\sqrt{x+x_{n}}\leq \sqrt{x+L}\leq L$, and you are done.
As a matter of fact please observe that the following relation is valid:
$\sqrt{x+L}=L$.
Thus, the sequence increases and is bounded therefore it converges. Now take the limit in the recurrence relation and the problem is solved.
I hope that there are no computational mistakes.
Didi

I don't know if you have shown that the limit of the recurrence is actually equal to the thing you calculated.  I mean an increasing sequence bounded from above necessarily converges, but to what limit?  How do you know you will get to the limit you caculated.  To  guarantee that, we have to show that we actually get closer and closer to the limit at a rate at which we can get within $\epsilon$ of it always.

I have shown that the sequence is bounded and monotone increasing ,therefore converges.

 Now, since the sequence verifies a recurrence relation, we are allowed to take the limit in the recurrence relation and thus you simply calculate the limit by solving an equation of degree 2.
Didi

Are you saying that recurrence relations can only converge to their fixed points if they converge?  That seems to be what you are assuming.

In this case it is not too hard to show that it converges to the particular fixed point at a rate at least geometrically so it will get there.

[quote="pdiao"]Are you saying that recurrence relations can only converge to their fixed points if they converge?[/quote]
That's exactly it. The general version:
Let $f$ be a continuous function from a topological space $X$ to itself. Let $x_{0}\in X$ be any point, and define $x_{1}=f(x_{0}),x_{2}=f(x_{1}),\dots,x_{n+1}=f(x_{n})$. If $x_{n}$ converges to $x$, then $f(x)=x$.
Proof: We have $x=\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}f(x_{n})= f\left(\lim_{n\to\infty}x_{n}\right)=f(x)$ by sequential continuity of $f$.


I'm interested to know, what exactly is the role of the monarch in a constitutional monarchy? For example in Great Britain, with the Prime Minister and Parliament and all, what power exactly does the queen have in government?

Also what power does she have in the Canadian government?

This gives a brief of what monarchs can do:

http://en.wikipedia.org/wiki/Constitutional_Monarchy

For Canada, I don't know.