$ \frac{1}{2}\geq a,b,c \geq 0$
$ a\plus{}b\plus{}c\equal{}1$
prove that
$ \frac{9}{32}\geq a^3\plus{}b^3\plus{}c^3\plus{}4abc$
 :lol:

Since $ a\plus{}b\plus{}c\equal{}1$ it follows that there exist the numbers $ x,y,z \ge 0$, $ x\plus{}y\plus{}z \not\equal{} 0$, such that:
$ a\equal{}\frac {x} {x\plus{}y\plus{}z}, b\equal{}\frac {y} {x\plus{}y\plus{}z}, c\equal{}\frac {z} {x\plus{}y\plus{}z}$
From $ \frac {1} {2} \ge a, b, c \ge 0$, we deduce that $ x, y, z$ are the sides of a triangle, maybe degenerate.
So, $ x\equal{}n\plus{}p$, $ y\equal{}m\plus{}p$, $ z\equal{}m\plus{}n$ with $ m, n, p \ge 0$ and $ m\plus{}n\plus{}p \not\equal{} 0$.
The inequality is equivalent with the following inequality:
$ \frac {9} {32}\ge\frac {(n\plus{}p)^3} {8(m\plus{}n\plus{}p)^3}\plus{}\frac {(m\plus{}p)^3} {8(m\plus{}n\plus{}p)^3}\plus{}\frac {(m\plus{}n)^3} {8(m\plus{}n\plus{}p)^3}\plus{}\frac {4(n\plus{}p)(m\plus{}p)(m\plus{}n)} {8(m\plus{}n\plus{}p)^3}$
After reductions, the inequality is equivalent with:
$ m^3\plus{}n^3\plus{}p^3\plus{}22mnp \ge m^2n\plus{}m^2p\plus{}n^2p\plus{}n^2m\plus{}p^2m\plus{}p^2n$

If $ m,n,p>0$ then the last one results from Schur:
$ m(m\minus{}n)(m\minus{}p)\plus{}n(n\minus{}m)(n\minus{}p)\plus{}p(p\minus{}m)(p\minus{}n) \ge 0$

If one of $ m,n,p$ is zero then the inequality is trivial:
$ m^3\plus{}n^3 \ge m^2n\plus{}mn^2 \Leftrightarrow (m\plus{}n)(m\minus{}n)^2 \ge 0$

The equality case is for $ p\equal{}0$ and $ m\equal{}n$ and the permutations.
In other words, when: $ (a,b,c)\equal{}\{ (\frac {1} {2}, \frac {1} {4}, \frac {1} {4}),(\frac {1} {4}, \frac {1} {2}, \frac {1} {4}),(\frac {1} {4}, \frac {1} {4}, \frac {1} {2}) \}$.


\footnote command are really helpful, but is troublesome if you have a great number in the same page.

I don’t know what kind of interactivity we can use in adobe, but I was thinking in something like “put your mouse on the Word and it will show a small box with the ‘footnote’”. But I think this kind of thing are not in the libraries of latex.

Just think in this kind of situation. You have 30 links in the same page going to another page with a explanation. You cannot make a footnote with this much of information. But for the reader will be troublesome (to not say boring) to go to the page, go back and make this process 30 times for each page!

I can make some solutions for this kind of problem. Like make a small explanation below each paragraph, make a lateral box.

I was just hoping that there were another’s solutions. You don’t need to make a complete code here, just give your idea, if it is sophisticated, you can put your code.

But the main point is that I don’t want to change PDF to another format.

The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=cooltooltips]cooltooltips package[/url] will do exactly this [i]put your mouse on the Word and it will show a small box with the ‘footnote’”.[/i]. The [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/cooltooltips/cooltooltips.pdf]documentation[/url] shows what's possible. Cooltooltips is supported by MiKTeX.

Really Cool. But...It do not make a PDF in LED but in anothers programs it do (TexMaker etc. (but with some errors))

Trying the code:

[code]\cooltooltip[0 0 1]{Example}{This is an example of a cool tooltip. Pretty cool, eh?}{http://www.ctan.org/}{Visit CTAN on the Web}{This text\strut}[/code]

Result from LED

[code]-----------------------
)
! Missing number, treated as zero.
<to be read again>
                   \pdfannot
l.4 ...g//}{Visit CTAN on the Web}{This text\strut}

?[/code]

eh... I think I will not use this package.

The old footnote were more simple, useful and old-fashioned.

It cannot depend on the editor unless you've misconfigured how it's compiled.
This complete minimal example works well:
[code]\documentclass{article}
\usepackage{cooltooltips}
\begin{document}
\cooltooltip[0 0 1]{Example}{This is an example of a cool tooltip. Pretty cool, eh?}{http://www.ctan.org/}{Visit CTAN on the Web}{This text\strut}
\end{document}[/code]and should work fine for you using pdflatex on any system.


geometry
circumcircle
geometric transformation
reflection
geometry unsolved

Let $ ABC$ be a triangle with $ A$ a right angle and $ \angle B<\angle C$. The tangent at $ A$ to the circumcircle $ k$ of $ \triangle ABC$ meets the line $ BC$ at $ D$. Let $ E$ be the reflection of $ A$ in $ BC$, $ X$ be the foot of the perpendicular from $ A$ to $ BE$, and $ Y$ the midpoint of segment $ AX$. Line $ BY$ intersect $ k$ again at $ Z$. Prove that $ BD$ is tangent to the circumcircle of $ \triangle ADZ$.

let $ AE,BC$ intersect at $ H$ and let $ AZ,BC$ intersect each other at point $ K$,so it sufficies to show that $ KD^2\equal{}KZ.KA$.

note that $ AZBE$ is cyclic so $ \angle ZBX\equal{}\angle ZAH$.

also in $ \triangle AEX$ points $ Y,H$ are the midpoints of sides $ AX,AE$ respectively so $ HY\parallel EX$ hence:

$ \angle HYB\equal{}\angle YBX\equal{}\angle ZAH$

thus $ YZAH$ is cyclic which means that $ \angle HYA\equal{}\angle HZA\equal{}90^\circ$.

but on the other hand we know that $ \angle AHB\equal{}90^\circ$ thus $ \angle ZHB\equal{}\angle ZAH$ therefore we have:

$ \triangle KZH\sim\triangle KAH\Rightarrow KH^2\equal{}KZ.KA\equal{}KB.KC$.

so in order to complete the proof,it's sufficient to show that $ KH\equal{}KD$,now let $ D'$ be the reflection of point $ H$ wrt point $ K$,now we have to show that $ D'\equiv D$.

but note that according to [b]newton formula[/b] we have $ \boxed{(D'H,BC)\equal{}\minus{}1} (I)$

on the other hand we know that $ \boxed{(DH,BC)\equal{}\minus{}1} (II)$

now according to (I),(II) we get that $ D\equiv D'$.

QED

seems that your problem is from shortlist 1998:
[url]http://www.mathlinks.ro/viewtopic.php?p=123012#123012[/url]

anyway here's another solution:

let $ G$ be a point on circle $ k$ such that $ AG$ is a diameter and let $ BD,AE$ intersect at $ H$.

first of all we prove that $ Z,H,G$ are collinear,note that we have:

$ \angle AEG \equal{} 90^\circ \equal{} \angle AXB$ and also $ \angle AGE \equal{} \angle ABX$ thus $ \triangle AGE\sim\triangle ABX$.

but on the other hand points $ H,Y$ are the midpoints of $ AE,AX$ respectively,hence $ \triangle AHG\sim\triangle AGE$ therefor:

$ \angle ABY \equal{} \angle AGH$ thus $ Z,H,G$ are collinear.

now in order to prove the tangency of $ BD$ and the circumcircle of $ \triangle AZD$,it's sufficient to show that $ \angle ZAD \equal{} \angle ZDC$.  (*)

now note that $ AD$ is tangent to circle $ k$ hence $ \angle AGZ \equal{} \angle ZAD$,thus for proving (*) we have to show that $ \angle AGZ \equal{} \angle ZDC$ or equivalently we have to prove that $ GOZD$ is cyclic:

$ \boxed{OH.HD \equal{} AH^2 \equal{} AH.HE} (I)$

$ \boxed{GH.HZ \equal{} AH.HE} (II)$

$ (I),(II)\Rightarrow OH.HD \equal{} GH.HZ$ thus $ GOZD$ is cyclic.

QED

See also
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=198944
Sincerely
Jean-Louis


inequalities
LaTeX
inequalities proposed

If $ a,b,c$ are nonnegative real numbers, then

$ \sum \frac {3a^2\minus{}bc}{2a^2\plus{}b^2\plus{}c^2} \le \frac 3{2}$.

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, then

$ \sum \frac {3a^2 \minus{} bc}{2a^2 \plus{} b^2 \plus{} c^2} \le \frac 3{2}$.[/quote]
After expansion we obtain:
$ \sum \frac {3a^2 \minus{} bc}{2a^2 \plus{} b^2 \plus{} c^2} \le \frac 3{2}\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(4a^5b \minus{} 9a^4b^2 \plus{} 5a^3b^3 \plus{} a^4bc \plus{} 6a^3b^2c \minus{} 7a^2b^2c^2)\geq0,$ which obviously true.
I have not found nothing without expansion. :(

I think there is at least a solution without expansion. :wink:

I found only the SOS form .It is easy to see that the inequality is equivalent to

$ \sum \frac{(a\minus{}b)^2(a^2\minus{}4ab\plus{}b^2\minus{}2ac\minus{}2bc\plus{}2c^2)}{(2a^2\plus{}b^2\plus{}c^2)(2b^2\plus{}c^2\plus{}a^2)}\leq 0$

But the $ 2a^2\plus{}b^2\plus{}c^2$ reminds me some problems of Vasc that solved with Cauchy Swartz .

For example $ \sum\frac{(b\plus{}c)^2}{2a^2\plus{}b^2\plus{}c^2}\leq 3$  :wink:

Yes, silouan. It can be rewritten as
\[ \frac12\sum_{\rm cyclic}\frac{(b\minus{}c)^2}{2a^2\plus{}b^2\plus{}c^2}\plus{}\sum_{\rm cyclic}\frac{(b\plus{}c)^2}{2a^2\plus{}b^2\plus{}c^2}\geq3.\]

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, then

$ \sum \frac {3a^2 \minus{} bc}{2a^2 \plus{} b^2 \plus{} c^2} \le \frac 3{2}$.[/quote]
$ \sum \frac {3a^2 \minus{} bc}{2a^2 \plus{} b^2 \plus{} c^2} \le \frac 3{2} \\ \Leftrightarrow \sum (\frac {3a^2 \minus{} bc}{2a^2 \plus{} b^2 \plus{} c^2} \minus{} \frac 1{2}) \le 0 \\ \Leftrightarrow \sum \frac {4a^2 \minus{}2 bc\minus{}b^2\minus{}c^2}{2a^2 \plus{} b^2 \plus{} c^2} \le 0 \\ \Leftrightarrow 2\sum \frac {b^2\plus{}c^2\minus{}2a^2}{2a^2 \plus{} b^2 \plus{} c^2} \ge \sum \frac {(b\minus{}c)^2}{2a^2 \plus{} b^2 \plus{} c^2}$
Notice that
$ \sum \frac {b^2\plus{}c^2\minus{}2a^2}{2a^2 \plus{} b^2 \plus{} c^2} \\ \equal{}\sum (a^2\minus{}b^2)(\frac{1}{2b^2\plus{}a^2\plus{}c^2}\minus{}\frac{1}{2a^2\plus{}b^2\plus{}c^2}) \\ \equal{} \sum \frac{(a^2\minus{}b^2)^2}{(2a^2\plus{}b^2\plus{}c^2)(2b^2\plus{}a^2\plus{}c^2)}$
the inequality is equivalent to
\[ \sum (a\minus{}b)^2 \left(\frac{2(a\plus{}b)^2}{(2a^2\plus{}b^2\plus{}c^2)(2b^2\plus{}a^2\plus{}c^2)}\minus{} \frac{1}{2c^2\plus{}a^2\plus{}b^2}\right) \ge 0
\]
This inequality can be rewritten as
\[ S_a(b\minus{}c)^2\plus{}S_b(c\minus{}a)^2\plus{}S_c(a\minus{}b)^2 \ge 0
\]
where
$ S_a\equal{}2(b\plus{}c)^2(2a^2\plus{}b^2\plus{}c^2)\minus{}(2b^2\plus{}a^2\plus{}c^2)(2c^2\plus{}a^2\plus{}b^2) \\ \equal{} a^2(b^2\plus{}c^2\plus{}8bc)\plus{}4bc(b^2\plus{}c^2)\minus{}b^2c^2\minus{}a^4 \\ \ge a^2b^2\minus{}a^4$
$ S_b \ge a^2b^2\minus{}b^4$
$ S_c \equal{}c^2(a^2\plus{}b^2\plus{}8ab)\plus{}4ab(a^2\plus{}b^2)\minus{}a^2b^2\minus{}c^4 \ge 0$
(we assume that $ a \ge b \ge c$)
It suffits to prove
$ a^2(b^2\minus{}a^2)(b\minus{}c)^2\plus{}b^2(a^2\minus{}b^2)(a\minus{}c)^2 \ge 0$
, which is clearly true because $ a \ge b \ge 0$ and $ ab\minus{}bc \ge ab\minus{}ac \ge 0$.
The inequality is proved.

Nice solution, ehku. :lol:
Another solution is as follows:
Since
$ S_a \equal{} 4bc(2a^2 \plus{} b^2 \plus{} c^2) \minus{} (a^2 \minus{} b^2)(a^2 \minus{} c^2)$,
we may write the inequality as
$ 4\sum bc(2a^2 \plus{} b^2 \plus{} c^2)(b \minus{} c)^2 \ge (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2$,
which is easy.

Much harder than this is the following:

$ \sum \frac {5a^2\plus{}4bc}{a^2\plus{}2b^2\plus{}2c^2} \ge \frac {27}{5}$,

because equality holds for (1,1,1) and (4,1,1).

[quote="Vasc"]Nice solution, ehku. :lol:

$ \sum \frac {5a^2 \plus{} 4bc}{a^2 \plus{} 2b^2 \plus{} 2c^2} \ge \frac {27}{5}$,

because equality holds for (1,1,1) and (4,1,1).[/quote]
Try for $ (a,b,c)\equal{}(1,0,0)$.

[quote="Vasc"][quote="ehku"][quote="Vasc"]Nice solution, ehku. :lol:

$ \sum \frac {5a^2 \plus{} 4bc}{a^2 \plus{} 2b^2 \plus{} 2c^2} \ge \frac {27}{5}$,

because equality holds for (1,1,1) and (4,1,1).[/quote]
Try for $ (a,b,c) \equal{} (1,0,0)$.[/quote]
Thanks, ehku. The reverse inequality holds; that is
$ \sum \frac {5a^2 \plus{} 4bc}{a^2 \plus{} 2b^2 \plus{} 2c^2} \le \frac {27}{5}$.

More general,

$ \sum \frac {p(3 \minus{} p)a^2 \plus{} 2(1 \minus{} p)bc}{pa^2 \plus{} b^2 \plus{} c^2} \le \frac {3(1 \plus{} p)(2 \minus{} p)}{2 \plus{} p}$,

for $ 0 < p < 1$ and any nonnegative $ a,b,c$. Equality holds for $ (1,1,1)$ and again for $ (\frac 2{p},1,1)$ or any cyclic permutation. For $ p > 1$, the reverse inequality holds.
I have no solution to these inequalities.  :oops:
For $ p \equal{} 2$ and $ p \equal{} 3$ we get two known inequalities.[/quote]
It is very interesting your inequality.
It seems be difficult to create and prove inequalities of this type. I have found some nice results, for example the following:

If $ a^2\plus{}b^2\plus{}c^2 \le 2(ab\plus{}bc\plus{}ca)$ then
\[ \frac{a^2\minus{}bc}{3a^2\plus{}b^2\plus{}c^2}\plus{}\frac{b^2\minus{}ca}{3b^2\plus{}c^2\plus{}a^2}\plus{}\frac{c^2\minus{}ab}{3c^2\plus{}a^2\plus{}b^2} \le 0
\]
For $ (a,b,c)\equal{}(1,1,0)$ or $ (1,1,1)$ or $ (1,1,4)$ it becomes equality.
Please try it, Vasc, there is a very interesting solution.
I am trying to find a general inequality for all above inequalities.

[quote="ehku"] If $ a^2 \plus{} b^2 \plus{} c^2 \le 2(ab \plus{} bc \plus{} ca)$ then
\[ \frac {a^2 \minus{} bc}{3a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {b^2 \minus{} ca}{3b^2 \plus{} c^2 \plus{} a^2} \plus{} \frac {c^2 \minus{} ab}{3c^2 \plus{} a^2 \plus{} b^2} \le 0
\]
For $ (a,b,c) \equal{} (1,1,0)$ or $ (1,1,1)$ or $ (1,1,4)$ it becomes equality. [/quote]
This is problem 18 in Algebraic Inequalities (p. 345 and 369), by Nguyen Anh Tuan.
Equality holds for $ a^2 \plus{} b^2 \plus{} c^2\equal{}2(ab \plus{} bc \plus{} ca)$.

As a consequence, in any triangle
\[ \sum \frac {a^4 \minus{} b^2c^2}{3a^4 \plus{} b^4 \plus{} c^4} \le 0.
\]

[quote="Vasc"][quote="ehku"] If $ a^2 \plus{} b^2 \plus{} c^2 \le 2(ab \plus{} bc \plus{} ca)$ then
\[ \frac {a^2 \minus{} bc}{3a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {b^2 \minus{} ca}{3b^2 \plus{} c^2 \plus{} a^2} \plus{} \frac {c^2 \minus{} ab}{3c^2 \plus{} a^2 \plus{} b^2} \le 0
\]
For $ (a,b,c) \equal{} (1,1,0)$ or $ (1,1,1)$ or $ (1,1,4)$ it becomes equality. [/quote]
This is problem 18 in Algebraic Inequalities (p. 345 and 369), by Nguyen Anh Tuan.
Equality holds for $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2(ab \plus{} bc \plus{} ca)$.

As a consequence, in any triangle
\[ \sum \frac {a^4 \minus{} b^2c^2}{3a^4 \plus{} b^4 \plus{} c^4} \le 0.
\]
[/quote]
 :o  Is that right ?
I hope the following identiy is not in your book:
For all reals $ a,b,c,p$ we have
\[ \frac{a^2\minus{}bc}{(p\plus{}1)a^2\plus{}b^2\plus{}c^2}\plus{}\frac{b^2\minus{}ca}{(p\plus{}1)b^2\plus{}c^2\plus{}a^2}\plus{}\frac{c^2\minus{}ab}{(p\plus{}1)c^2\plus{}a^2\plus{}b^2} \\ \equal{}\frac{(\sum a^2\minus{}p\sum ab)[(\sum a^2)(\sum a^2\minus{}\sum ab)\plus{}p(\sum a^2b^2\minus{}abc \sum a)]}{((p\plus{}1)a^2\plus{}b^2\plus{}c^2)((p\plus{}1)b^2\plus{}c^2\plus{}a^2)((p\plus{}1)c^2\plus{}a^2\plus{}b^2)}
\]

[quote="Vasc"][quote="ehku"][quote="Vasc"]Nice solution, ehku. :lol:

$ \sum \frac {5a^2 \plus{} 4bc}{a^2 \plus{} 2b^2 \plus{} 2c^2} \ge \frac {27}{5}$,

because equality holds for (1,1,1) and (4,1,1).[/quote]
Try for $ (a,b,c) \equal{} (1,0,0)$.[/quote]
Thanks, ehku. The reverse inequality holds; that is
$ \sum \frac {5a^2 \plus{} 4bc}{a^2 \plus{} 2b^2 \plus{} 2c^2} \le \frac {27}{5}$.

More general,

$ \sum \frac {p(3 \minus{} p)a^2 \plus{} 2(1 \minus{} p)bc}{pa^2 \plus{} b^2 \plus{} c^2} \le \frac {3(1 \plus{} p)(2 \minus{} p)}{2 \plus{} p}$,

for $ 0 < p < 1$ and any nonnegative $ a,b,c$. Equality holds for $ (1,1,1)$ and again for $ (\frac 2{p},1,1)$ or any cyclic permutation. For $ p > 1$, the reverse inequality holds.
I have no solution to these inequalities.  :oops:
For $ p \equal{} 2$ and $ p \equal{} 3$ we get two known inequalities.[/quote]
This inequality is true for all reals $ a,b,c$.

I have proved it in the small case $ \frac{1}{2} \le p \le 1$ (that also means the inequality $ \sum \frac {5a^2 \plus{} 4bc}{a^2 \plus{} 2b^2 \plus{} 2c^2} \le \frac {27}{5}$ was proved, of course, if I didn't make any mistake).
To prove the original inequality in the case $ 0 \le p \le 1$, it reduces to prove
\[ (pb^2\plus{}c^2\plus{}a^2)(pa\minus{}2c)^2\plus{}(pc^2\plus{}a^2\plus{}b^2)(pa\minus{}2b)^2 \ge (1\minus{}p)(p\plus{}2)(b^2\minus{}c^2)^2
\]
for all $ a \ge b\ge c \ge 0$.
I think it is true  for all $ 1 \ge p \ge 0$ but I can only prove it for $ \frac{1}{2} \le p \le 1$.

The case $ p \ge 1$ is much more difficult.

[quote="manlio"][quote="Vasc"]
Another solution is as follows:
Since
$ S_a \equal{} 4bc(2a^2 \plus{} b^2 \plus{} c^2) \minus{} (a^2 \minus{} b^2)(a^2 \minus{} c^2)$,
we may write the inequality as
$ 4\sum bc(2a^2 \plus{} b^2 \plus{} c^2)(b \minus{} c)^2 \ge (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2$,
which is easy.
[/quote]

Vasc, why do you say it is easy? How do you finish your proof?

Thank you very much.[/quote]
Well, to prove this one, we will asume $ a \ge b \ge c$, then we have
\[ 4\sum bc(2a^2\plus{}b^2\plus{}c^2)(b\minus{}c)^2 \ge 4ab(a^2\plus{}b^2\plus{}2c^2) (a\minus{}b)^2 \ge  4ab(a^2\plus{}b^2)(a\minus{}b)^2 \ge 8a^2b^2(a\minus{}b)^2
\]
Moreover,
\[ (a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2 \le a^2b^2(a\minus{}b)^2
\]
hence the inequality is true. :)

[quote="can_hang2007"][quote="manlio"][quote="Vasc"]
Another solution is as follows:
Since
$ S_a \equal{} 4bc(2a^2 \plus{} b^2 \plus{} c^2) \minus{} (a^2 \minus{} b^2)(a^2 \minus{} c^2)$,
we may write the inequality as
$ 4\sum bc(2a^2 \plus{} b^2 \plus{} c^2)(b \minus{} c)^2 \ge (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2$,
which is easy.
[/quote]

Vasc, why do you say it is easy? How do you finish your proof?

Thank you very much.[/quote]
Well, to prove this one, we will asume $ a \ge b \ge c$, then we have
\[ 4\sum bc(2a^2 \plus{} b^2 \plus{} c^2)(b \minus{} c)^2 \ge 4ab(a^2 \plus{} b^2 \plus{} 2c^2) (a \minus{} b)^2 \ge 4ab(a^2 \plus{} b^2)(a \minus{} b)^2 \ge 8a^2b^2(a \minus{} b)^2
\]
Moreover,
\[ (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \le a^2b^2(a \minus{} b)^2
\]
hence the inequality is true. :)[/quote]

Very nice and very simple :)

Thank you very much.

[quote="ehku"][/quote][quote="Vasc"][quote="ehku"][quote="Vasc"]Nice solution, ehku. :lol:

$ \sum \frac {5a^2 + 4bc}{a^2 + 2b^2 + 2c^2} \ge \frac {27}{5}$,

because equality holds for (1,1,1) and (4,1,1).[/quote]
Try for $ (a,b,c) = (1,0,0)$.[/quote]
Thanks, ehku. The reverse inequality holds; that is
$ \sum \frac {5a^2 + 4bc}{a^2 + 2b^2 + 2c^2} \le \frac {27}{5}$.

More general,

$ \sum \frac {p(3 - p)a^2 + 2(1 - p)bc}{pa^2 + b^2 + c^2} \le \frac {3(1 + p)(2 - p)}{2 + p}$,

for $ 0 < p < 1$ and any nonnegative $ a,b,c$. Equality holds for $ (1,1,1)$ and again for $ (\frac 2{p},1,1)$ or any cyclic permutation. For $ p > 1$, the reverse inequality holds.
I have no solution to these inequalities.  :oops:
For $ p = 2$ and $ p = 3$ we get two known inequalities.[/quote][quote="ehku"]
This inequality is true for all reals $ a,b,c$.

I have proved it in the small case $ \frac {1}{2} \le p \le 1$ (that also means the inequality $ \sum \frac {5a^2 + 4bc}{a^2 + 2b^2 + 2c^2} \le \frac {27}{5}$ was proved, of course, if I didn't make any mistake).
To prove the original inequality in the case $ 0 \le p \le 1$, it reduces to prove
\[ (pb^2 + c^2 + a^2)(pa - 2c)^2 + (pc^2 + a^2 + b^2)(pa - 2b)^2 \ge (1 - p)(p + 2)(b^2 - c^2)^2
\]
for all $ a \ge b\ge c \ge 0$.
I think it is true  for all $ 1 \ge p \ge 0$ but I can only prove it for $ \frac {1}{2} \le p \le 1$.

The case $ p \ge 1$ is much more difficult.[/quote]
It was proved for all $ p \ge 3$.
The useful lemma posted here http://www.mathlinks.ro/viewtopic.php?p=1042081#1042081 can help us to prove it for $ 0 < p < \frac {1}{2}$ or $ 1 < p < 3$, then the above inequality can be proved completely.
I will try prove two these cases by the same way for $ 1 \ge p \ge \frac {1}{2}$ or $ p \ge 3$.

[quote="ehku"] The useful lemma posted here http://www.mathlinks.ro/viewtopic.php?p=1042081#1042081 can help us to prove it [/quote]
This lemma is a particular case of EV-Theorem. For this case, the method is too laborious.

Some ideas for Mathlinkers.
This is an inequality of six degree. There is an inequality of four degree with the same equality conditions:

$ p^2\sum a^4 \plus{} 4(p \plus{} 1)\sum a^2b^2 \plus{} (p^2 \minus{} 4)\sum a^2bc \minus{} (p^2 \plus{} 2p)\sum ab(a^2 \plus{} b^2) \ge 0$,

which is equivalent to $ \frac 1{2}\sum(pb \plus{} pc \minus{} (p \plus{} 2)a)^2(b \minus{} c)^2 \ge 0$.
Can we use this to prove our difficult inequality ?

The inequality becomes identity for $ p \equal{} 1$. Dividing by $ p \minus{} 1$ :o , we find an inequality for any $ p > 0$. It is interesting to prove this inequality.

The inequality was proved for all $ p \ge \frac{1}{2}$ and all reals $ a,b,c$.

[quote]This lemma is a particular case of EV-Theorem. For this case, the method is too laborious.[/quote]
Maybe it is not nice, but it is not laborious, because we needn't expand it.

[hide="Dear Vasc"][quote="ehku"]
I hope the following identiy is not in your book:
For all reals $ a,b,c,p$ we have
\[ \frac {a^2 \minus{} bc}{(p \plus{} 1)a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {b^2 \minus{} ca}{(p \plus{} 1)b^2 \plus{} c^2 \plus{} a^2} \plus{} \frac {c^2 \minus{} ab}{(p \plus{} 1)c^2 \plus{} a^2 \plus{} b^2} \\
\equal{} \frac {(\sum a^2 \minus{} p\sum ab)[(\sum a^2)(\sum a^2 \minus{} \sum ab) \plus{} p(\sum a^2b^2 \minus{} abc \sum a)]}{((p \plus{} 1)a^2 \plus{} b^2 \plus{} c^2)((p \plus{} 1)b^2 \plus{} c^2 \plus{} a^2)((p \plus{} 1)c^2 \plus{} a^2 \plus{} b^2)}
\]
[/quote]
 :blush:

The inequality
\[ \frac {a^2 \minus{} bc}{3a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {b^2 \minus{} ca}{3b^2 \plus{} c^2 \plus{} a^2} \plus{} \frac {c^2 \minus{} ab}{3c^2 \plus{} a^2 \plus{} b^2} \le 0
\]
for $ a^2 \plus{} b^2 \plus{} c^2 \le 2(ab \plus{} bc \plus{} ca)$ can be proved without the above indentity:
We need only consider two case $ a \ge b \ge c \ge 0$ or $ a\ge b \ge 0 > c$.
 If $ a\ge b \ge 0 \ge c$ then $ a^2 \plus{} b^2 \plus{} c^2 \le 2(ab \plus{} bc \plus{} ca) \le 2ab$, contradiction. Then $ a \ge b \ge c \ge 0$.
Because
\[ (a^2 \minus{} bc)(b\plus{}c) \ge (b^2 \minus{} ca)(c\plus{}a) \ge (c^2 \minus{} ab)(a\plus{}b) \\ (3a^2 \plus{} b^2 \plus{} c^2)(b\plus{}c) \ge (3b^2 \plus{} c^2 \plus{}a^2)(c\plus{}a) \ge (3c^2 \plus{} a^2 \plus{} b^2)(a\plus{}b) >0
\]
, by Chebysev inequality we get
\[ LHS \le (\sum (a^2 \minus{} bc)(b\plus{}c))\left(\sum \frac{1}{(3a^2 \plus{} b^2 \plus{} c^2)(b\plus{}c)}\right)\equal{}0
\]
because $ \sum (a^2 \minus{} bc)(b\plus{}c)\equal{}0$.[/hide]

...and now, it was proved for all positive real $ p$ ! :!:

[b]Problem.[/b] Let $ a,b,c,p$ be real numbers. We denote
\[ F(a,b,c)\equal{}\sum \frac {p(3 \minus{} p)a^2 \plus{} 2(1 \minus{} p)bc}{pa^2 \plus{} b^2 \plus{} c^2} \minus{} \frac {3(1 \plus{} p)(2 \minus{} p)}{2 \plus{} p}
\].
Prove that $ (p\minus{}1).F(a,b,c) \ge 0$ for all real numbers $ a,b,c$ and all positive real number $ p$ with equality if and only if $ p\equal{}1$ or $ (a,b,c)\equal{}(1,1,1)$ or $ (a,b,c)\equal{}(1,1,\frac{2}{p})$ and their permutations.
[b]Solution.[/b]
Because \[ (2\plus{}p)[p(3\minus{}p)a^2\plus{}2(1\minus{}p)bc]\minus{}(1\plus{}p)(2\minus{}p)(pa^2\plus{}b^2\plus{}c^2) \\ \equal{}2p(2a^2\minus{}b^2\minus{}c^2)\plus{}(p\minus{}1)(p\plus{}2)(b\minus{}c)^2
\]
, we have
\[ F(a,b,c)\equal{}\sum \left(\frac {p(3 \minus{} p)a^2 \plus{} 2(1 \minus{} p)bc}{pa^2 \plus{} b^2 \plus{} c^2}\minus{}\frac {(1 \plus{} p)(2 \minus{} p)}{2 \plus{} p}\right) \\ \equal{} \frac{2p}{2\plus{}p} \sum \frac{2a^2\minus{}b^2\minus{}c^2}{pa^2\plus{}b^2\plus{}c^2}\plus{}(p\minus{}1)\sum \frac{(b\minus{}c)^2}{pa^2\plus{}b^2\plus{}c^2}.
\]
Notice that
\[ \sum \frac{2a^2\minus{}b^2\minus{}c^2}{pa^2\plus{}b^2\plus{}c^2} \equal{} \sum(a^2\minus{}b^2)\left(\frac {1}{pa^2 \plus{} b^2 \plus{} c^2}\minus{}\frac {1}{pb^2\plus{}c^2\plus{}a^2}\right) \\ \equal{}(1\minus{}p) \sum \frac{(a^2\minus{}b^2)^2}{(pa^2 \plus{} b^2 \plus{} c^2)(pb^2\plus{}c^2\plus{}a^2)}
\]
, we have
\[ F(a,b,c)\equal{}\frac{2p(1\minus{}p)}{2\plus{}p}\sum \frac{(a^2\minus{}b^2)^2}{(pa^2 \plus{} b^2 \plus{} c^2)(pb^2\plus{}c^2\plus{}a^2)}\plus{} \\ \plus{}(p\minus{}1)\sum \frac{(b\minus{}c)^2}{pa^2\plus{}b^2\plus{}c^2} \\ \equal{} (p\minus{}1) \left(\sum \frac{(a\minus{}b)^2}{pc^2\plus{}a^2\plus{}b^2} \minus{}\frac{2p}{2\plus{}p} \sum \frac{(a^2\minus{}b^2)^2}{(pa^2 \plus{} b^2 \plus{} c^2)(pb^2\plus{}c^2\plus{}a^2)}\right).
\]
Hence the original inequality is equivalent to
\[ \sum \frac{(a\minus{}b)^2}{pc^2\plus{}a^2\plus{}b^2} \ge \frac{2p}{2\plus{}p} \sum \frac{(a^2\minus{}b^2)^2}{(pa^2 \plus{} b^2 \plus{} c^2)(pb^2\plus{}c^2\plus{}a^2)}
\]
for all reals $ a,b,c$ and positive real $ p$.
From the inequality $ (x\minus{}y)^2 \ge (|x|\minus{}|y|)^2 \ \forall x,y$, we see that it suffits to prove the above inequality for $ a,b,c \ge 0$ and $ p > 0$.
This inequality is equivalent to
\[ \sum (a\minus{}b)^2\left(\frac{2\plus{}p}{pc^2\plus{}a^2\plus{}b^2} \minus{}\frac{2p(a\plus{}b)^2}{(pa^2 \plus{} b^2 \plus{} c^2)(pb^2\plus{}c^2\plus{}a^2)}\right) \ge 0.
\]
, which can be rewritten as
\[ S_a(b\minus{}c)^2\plus{}S_b(c\minus{}a)^2\plus{}S_c(a\minus{}b)^2 \ge 0
\]
with
\[ S_c\equal{}(p\plus{}2)(pa^2 \plus{} b^2 \plus{} c^2)(pb^2\plus{}c^2\plus{}a^2)\minus{}2p(a\plus{}b)^2(pc^2\plus{}a^2\plus{}b^2)
\]
and $ S_a, S_b$ are determined similarly.
We can rewritten $ S_c$ as two forms
\[ S_c\equal{}(pc^2\plus{}a^2\plus{}b^2)(pb\minus{}2a)^2\plus{}(p\plus{}2)(c^2\minus{}a^2)[pc^2\plus{}a^2\plus{}b^2\plus{}(1\minus{}p)(pb^2\plus{}a^2\plus{}c^2)] \\ S_c\equal{}(pc^2\plus{}a^2\plus{}b^2)(pa\minus{}2b)^2\plus{}(p\plus{}2)(c^2\minus{}b^2)[pc^2\plus{}a^2\plus{}b^2\plus{}(1\minus{}p)(pa^2\plus{}b^2\plus{}c^2)]
\]
and similarly for $ S_a$ and $ S_b$.

It is very useful we rewrite in both two above forms.

(see my next post).

Now, we assume $ a \ge b \ge c \ge 0$ and consider three cases of $ p$:

[b]First case : $ 0 <p \le 1$.[/b]
Clearly that $ S_a \ge 0, S_b \ge 0$, it suffits to prove $ S_b\plus{}S_c \ge 0$ (because $ a\minus{}c \ge a\minus{}b \ge 0$).
We have
\[ S_b\plus{}S_c\equal{}(pb^2\plus{}c^2\plus{}a^2)(pa\minus{}2c)^2\plus{}(pc^2\plus{}a^2\plus{}b^2)(pa\minus{}2b)^2\minus{}(p\plus{}2)(1\minus{}p)(b^2\minus{}c^2)^2.
\]
Using the inequality $ 2(x^2\plus{}y^2) \ge (x\plus{}y)^2 \ \forall x,y$ we have
\[ (pb^2\plus{}c^2\plus{}a^2)(pa\minus{}2c)^2\plus{}(pc^2\plus{}a^2\plus{}b^2)(pa\minus{}2b)^2 \\ \equal{}(a^2\plus{}pb^2\plus{}pc^2)((pa\minus{}2b)^2\plus{}(pa\minus{}2c)^2)\plus{}(1\minus{}p)((pab\minus{}2b^2)^2\plus{}(pac\minus{}2c^2)^2) \\ \ge (a^2\plus{}pb^2\plus{}pc^2)\frac{1}{2}(2b\minus{}2c)^2\plus{}(1\minus{}p)\frac{1}{2}(2b^2\minus{}2c^2\plus{}pac\minus{}pbc)^2 \\ \ge (a^2\plus{}\frac{1}{2}(b\plus{}c)^2).2.(b\minus{}c)^2\plus{}\frac{1}{2}(1\minus{}p)(b\minus{}c)^2(2b\plus{}2c\minus{}pa)^2.
\]
We need prove
\[ 2a^2\plus{}p(b\plus{}c)^2\plus{}\frac{1}{2}(1\minus{}p)(2b\plus{}2c\minus{}pa)^2 \ge (2\plus{}p)(1\minus{}p)(b\plus{}c)^2
\]
or
\[ (4\plus{}p^2\minus{}p^3)a^2\plus{}2p^2(b\plus{}c)^2 \ge 4p(1\minus{}p)a(b\plus{}c)
\]
, which is true because $ (4\plus{}p^2\minus{}p^3)a^2\plus{}2p^2(b\plus{}c)^2 \ge 4a^2 \plus{}2p^2(b\plus{}c)^2 \ge 4p(1\minus{}p)a(b\plus{}c)$.
The inequality was proved for $ 0<p \le 1$.

[b]Second case: $ 1 < p \le \frac{3}{2}$.[/b]
From the expression of $ S_b\plus{}S_c$ in the first case, we see that $ S_a\plus{}S_b \ge 0,S_b\plus{}S_c \ge 0,S_c\plus{}S_a \ge 0$ for $ 1 < p$. It suffits to prove $ S_b \ge 0$, which is true because $ b \ge c$ and
\[ pb^2\plus{}c^2\plus{}a^2\plus{}(1\minus{}p)(pa^2\plus{}b^2\plus{}c^2)\equal{}b^2\plus{}(2\minus{}p)c^2\plus{}(1\plus{}p\minus{}p^2)a^2 \ge b^2\plus{}(1\plus{}p\minus{}p^2)b^2 \ge 0.
\]
The inequality was proved in this case.

[b]Third case: $ \frac{3}{2} \le p$.[/b]
For this case, we rewrite again as
\[ S_c\equal{}(pc^2\plus{}a^2\plus{}b^2)(pb\minus{}2a)^2\plus{}(2\plus{}p)(c^2\minus{}a^2)[c^2\minus{}b^2\plus{}(2\minus{}p)(a^2\plus{}b^2\plus{}pb^2)] \\ \equal{}K_c\plus{}(2\plus{}p)(c^2\minus{}a^2)(c^2\minus{}b^2)
\]
with
\[ K_c\equal{}(pc^2\plus{}a^2\plus{}b^2)(pb\minus{}2a)^2\plus{}(4\minus{}p^2)(c^2\minus{}a^2)(a^2\plus{}b^2\plus{}pb^2).
\]
We also have another form of $ K_c$ as
\[ K_c\equal{}(pc^2\plus{}a^2\plus{}b^2)(pa\minus{}2b)^2\plus{}(4\minus{}p^2)(c^2\minus{}b^2)(b^2\plus{}a^2\plus{}pa^2).
\]
(It is similar for $ K_a$ and $ K_b$)
Then
\[ \sum S_c (a\minus{}b)^2\equal{}\sum K_c(a\minus{}b)^2\plus{}\sum (2\plus{}p)(c^2\minus{}a^2)(c^2\minus{}b^2)(a\minus{}b)^2.
\]
Because
\[ \sum (c^2\minus{}a^2)(c^2\minus{}b^2)(a\minus{}b)^2\equal{}(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2 \ge 0
\]
, we need prove
\[ K_c(a\minus{}b)^2\plus{}K_b(a\minus{}c)^2\plus{}K_a(b\minus{}c)^2 \ge 0.
\]
Clearly that $ K_b$ is always nonnegative (from $ a \ge b \ge c$) and $ K_a \ge 0$ for $ p\le 2$, $ K_c \ge 0$ for $ p\ge 2$ (where we used both two forms of them).
We have
\[ K_b\plus{}K_c\equal{}(pb^2\plus{}c^2\plus{}a^2)(pa\minus{}2c)^2\plus{}(pc^2\plus{}a^2\plus{}b^2)(pa\minus{}2b)^2\minus{}(4\minus{}p^2)(b^2\minus{}c^2)^2 \\ K_b\plus{}K_a\equal{}(pb^2\plus{}c^2\plus{}a^2)(pc\minus{}2a)^2\plus{}(pa^2\plus{}c^2\plus{}b^2)(pc\minus{}2b)^2\minus{}(4\minus{}p^2)(b^2\minus{}a^2)^2
\]
From this, if $ p \ge 2$ then the inequality is clearly true.
If $ 2 \ge p \ge \frac{3}{2}$, then $ K_a \ge 0$ and we must prove $ K_b\plus{}K_c \ge 0$, which is true because
\[ (pb^2\plus{}c^2\plus{}a^2)(pa\minus{}2c)^2\plus{}(pc^2\plus{}a^2\plus{}b^2)(pa\minus{}2b)^2 \\ \equal{} \frac{1}{2}(p\plus{}2)(b^2\plus{}c^2)(pa\minus{}2c)^2\plus{}\frac{1}{2}(p\plus{}2)(b^2\plus{}c^2)(pa\minus{}2b)^2 \\ \equal{} \frac{1}{2}(p\plus{}2)(b^2\plus{}c^2)((pa\minus{}2c)^2\plus{}(pa\minus{}2b)^2) \\ \ge \frac{1}{2}(p\plus{}2)(b^2\plus{}c^2).\frac{1}{2}(2b\minus{}2c)^2 \\ \equal{}(p\plus{}2)(b^2\plus{}c^2)(b\minus{}c)^2 \\ \ge (4\minus{}p^2)(b^2\minus{}c^2)^2.
\]
The inequality was proved in the last case.

It is my last solution in this forum :(  (then I typed Latex carefully  :lol: , sorry if there is some mistakes). Good bye Vasc and all!

Your complete solution is magnificent, ehku! Congratulations. :lol:

The key of your solution is to use the idea from my solution to the initial proposed inequality, to simplify the form of $ S_a, S_b, S_c$ by means of the expression $ (a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2$. You have used this idea in a brilliant solution.

I found a complete solution to this inequality by EV-Theorem.

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, then

$ \sum \frac {3a^2 \minus{} bc}{2a^2 \plus{} b^2 \plus{} c^2} \le \frac 3{2}$.[/quote]
My solution is the following
:)

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, then

$ \sum \frac {3a^2 \minus{} bc}{2a^2 \plus{} b^2 \plus{} c^2} \le \frac 3{2}$.[/quote]

This inequality is presented with three nice solutions (see [b]Problem 2.92[/b], page 144) in new book:

Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://canhang2007.wordpress.com/2009/12/08/new-book-inequalities-with-beautiful-solutions-v-cirtoaje-v-q-b-can-t-q-anh/][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house 2009.


calculus
integration
logarithms
calculus computations

maybe i have asked this before...  does anyone here know how to solve these 2 integrals (using some clever substitution) ?  :maybe:

$\textbf{(i)}\;\;\;\int_{0}^\infty \frac{\ln^{3}{(1+x^{2})}}{x^{2}}\;dx$

$\textbf{(ii)}\;\;\;\int_{0}^\infty \frac{\ln^{3}{(1+x^{2})}}{x^{4}}\;dx$

I think you should do this:

1) in tegrate by parts and you get that

$\int_{0}^{\infty}\frac{\ln^{3}(1+x^{2})}{x^{3}}dx=6\int_{0}^{\infty}\frac{\ln^{2}(1+x^{2})}{1+x^{2}}dx=24\int_{0}^{\pi/2}\ln^{2}\cos(t)dx$, and now you know how to calculate this integral.

2) Integrate by parts and use simple fraction decomposition and you have:

$I=\int_{0}^{\infty}\frac{\ln^{3}(1+x^{2})}{x^{3}}dx=2\left(\int_{0}^{\infty}\frac{\ln^{2}(1+x^{2})}{x^{2}}dx-\int_{0}^{\infty}\frac{\ln^{2}(1+x^{2})}{1+x^{2}}dx\right)$

The first integral can be calculated by parts one more time and you get that

$I=2\left(4\int_{0}^{\infty}\frac{\ln(1+x^{2})}{1+x^{2}}dx-\int_{0}^{\infty}\frac{\ln^{2}(1+x^{2})}{1+x^{2}}dx\right)$

$I=2\left(-8\int_{0}^{\pi/2}\ln\cos(t)dt-4\int_{0}^{\pi/2}\ln^{2}\cos(t)dt\right)$.

Therefore you need to calculate the following two integrals;

$\int_{0}^{\pi/2}\ln^{2}\cos(t)dt$

and

$\int_{0}^{\pi/2}\ln(\cos(t))dt$.

I hope I made no computational mistakes.

[quote="didilica"]I think you should do this:

1) Integrate by parts and you get that

$I=\int_{0}^{\infty}\frac{\ln^{3}(1+x^{2})}{x^{2}}dx=6\int_{0}^{\infty}\frac{\ln^{2}(1+x^{2})}{1+x^{2}}dx=24\int_{0}^{\pi/2}\ln^{2}\cos t\;dt$,
and now you know how to calculate this integral.

2) Integrate by parts and use simple fraction decomposition and you have:

$I=\int_{0}^{\infty}\frac{\ln^{3}(1+x^{2})}{x^{4}}dx=2\left(\int_{0}^{\infty}\frac{\ln^{2}(1+x^{2})}{x^{2}}dx-\int_{0}^{\infty}\frac{\ln^{2}(1+x^{2})}{1+x^{2}}dx\right)$

...
...
...
[/quote]

just wanted to make 2 minor corrections (power of $x$ in the
denominator of the 2 integrals)  :|


calculus
integration
function
trigonometry
calculus computations

Find the condition of $a,b$ for which the function $f(x)\ (0\leq x\leq 2\pi)$ satisfying the following equality can be determined uniquely,then determine $f(x)$, assuming that $f(x) $ is a continuous function at $0\leq x\leq 2\pi$.

\[f(x)=\frac{a}{2\pi}\int_0^{2\pi} \sin (x+y)f(y)dy+\frac{b}{2\pi}\int_0^{2\pi} \cos (x-y)f(y)dy+\sin x+\cos x\]

I'm not too sure about the restrictions on $a$ and $b$, but here's what I've come up with:

$f(x)=\frac{a}{2\pi}\int_0^{2\pi} \sin (x+y)f(y)dy+\frac{b}{2\pi}\int_0^{2\pi} \cos (x-y)f(y)dy+\sin x+\cos x$

$=\left(\frac{a}{2\pi}\sin x + \frac{b}{2\pi} \cos x\right)\int_0^{2\pi} \cos y f(y) \ dy + \left(\frac{b}{2\pi}\sin x + \frac{a}{2\pi} \cos x \right)\int_0^{2\pi} \sin y f(y) \ dy$ $ + \sin x + \cos x$

Since $\int_0^{2\pi} \cos y f(y) \ dy$ and $\int_0^{2\pi} \sin y f(y) \ dy$ are constants we can call them $\alpha$ and $\beta$ respectively, and now we have:

$\alpha = \int_0^{2\pi} \cos y f(y) \ dy = \frac{a\beta+b\alpha+2\pi}{2}\Rightarrow \alpha = \frac{2\pi+a\beta}{2-b}$

and

$\beta =  \int_0^{2\pi} \sin y f(y) \ dy = \frac{a\alpha+b\beta+2\pi}{2}\Rightarrow \beta = \frac{2\pi+a\alpha}{2-b}$

Solving this system for $(\alpha, \beta)$ (I think this is where the restrictions on $a$ and $b$ come in), gives:

[b]Case 1:[/b]

$a\not= 0$, $b=a+2$, $(\alpha,\beta) = \left(\frac{-aC_1+2\pi}{a},C_1\right)$  Obviously this is not unique.

[b]Case 2:[/b]

$a+b\not=2$, $(\alpha, \beta) = \left(-\frac{2\pi}{a+b-2}, -\frac{2\pi}{a+b-2}\right)$

So finally we have

$f(x) = \left[\left(\frac{a}{2\pi} \sin x+ \frac{b}{2\pi} \cos x \right)+\left(\frac{b}{2\pi} \sin x+\frac{a}{2\pi} \cos x \right) \right]\cdot\left(-\frac{2\pi}{a+b-2}\right)$

$= -\frac{2(\cos x + \sin x)}{a+b-2}$

Thank you for your reply, Jimmy.

You can use the fact the system of equation $px+qy=0$ and $rx+sy=0$ have a pair of $(x,y)$ uniquely $\Longleftrightarrow ps-qr\neq 0$.

P.S. Your answer for $f(x)$ is correct. :)

kunny

I forgot about Cramer's rule.   :)

So, from the equations

$\alpha  - \frac{a\beta}{2-b} = \frac{2\pi}{2-b}$

and

$\beta - \frac{a\alpha}{2-b} = \frac{2\pi}{2-b}$

we have $1-\left(\frac{a}{2-b}\right)^2 \not=0\Rightarrow a+b\not=2 \text{ and } b\not=2$

Let me know if I did this right, Thanks.


[hide]Break up into 1; 2, 1; 3, 2, 1; etc.

There is one term in the first group, two terms in the second group, and the first number in group n is n.

1+2+3+4+..+13=91. The 92nd term is 14. That means the 100th term is 6.  [/hide]


$\{a_n\}$ is an arithmetic sequence whose first term and common difference are positive. If $a_3$, $a_5$ and $a_k$ form a geometric sequence and $a_9=40$, find the expression for $a_n$.

Maybe you could use Soni's Theorem...or maybe not...

What's Soni's Theorem?

Anyway, you only need to know the basic stuff about arithmetic sequence and geometric sequence.

[hide="Hint"]
Express $d$ (common difference of the arithmetic sequence) in terms of $k$.
Then it turns out that there's only one possible value for $k$...
[/hide]


[quote="shobber"][quote="warut_suk"]What are $a,b,c$?[/quote]
He means:
$a+b+c=3$
$a^2+b^2+c^2=5$
$a^3+b^3+c^3=7$.[/quote]

Oh, I'm really sorry for being careless.

The following identity is why:

$a^n+b^n+c^n = (a+b+c)(a^{n-1}+b^{n-1}+c^{n-1})-(ab+ac+bc)(a^{n-2}+b^{n-2}+c^{n-2}) + abc(a^{n-3}+b^{n-3}+c^{n-3})$

Generalized, with $k_i$ being the $i$th symmetric sum for $x_1,...,x_j$ and $S_n$ the $n$th power sum

$S_{n+1}=k_1S_n + k_2S_{n-1}+...$