[/quote]
,
,
and 
are equal, and they intersect at 
.
and 
intersect at point 
lying inside triangle 
,
and 
intersect at point 
lying inside triangle 
.
.
denote the set of positive rational numbers. Find, with proof, all functions 
such that, for all positive rational numbers 
and 
we have ![\[f(x)=f(x+y)+f(x+x^2f(y)).\]](http://latex.artofproblemsolving.com/9/e/a/9ea71d2ee277dc3c689ba2d791b451e29353edb3.png)
be a triangle with circumcenter 
.
be the intersections of the perpendicular bisectors of 
and 
.
and 
and the point 
is given. Let 
be set of all 
segments of real line of type ![$[i, j]$](http://latex.artofproblemsolving.com/a/6/4/a644edd67caa59eaded765c417b6d9822f782ec2.png)
,
and 
are integers, 
.
is said to be valuable if the intersection of any two segments from 
is either empty, or is a segment of nonzero length belonging to 
rows of coins. There is 1 coin in the first row, 2 coins in the second row, 3 coins in the third row, and so on, up to 
satisfying 
,![\[ \frac{a^4+5b^4}{a(a+2b)} + \frac{b^4+5c^4}{b(b+2c)} + \frac{c^4+5a^4}{c(c+2a)} \geq 1- ab-bc-ca \]](http://latex.artofproblemsolving.com/f/e/8/fe85cf52b883c214806ab837a2eb25cebb511d7b.png)
such that there exists a unique integer 
such that 
with the following property:![\[ n!\mid a^n + 1
\]](http://latex.artofproblemsolving.com/e/2/3/e23d39aab685888993d355d8c93d4e6475f53487.png)
(use in a problem for junior) to practice. Can anyone share to me?
side length and with center 
.
is drawn where point 
.
be an odd integer, 
.
integers 
such that for any 
,
and 
.
for which satisfy 
for some 
,
is greater than 
.
such that![\[f(x+yf(x))+f(xf(y)-y)=f(x)-f(y)+2xy\]](http://latex.artofproblemsolving.com/e/b/c/ebccae7e765d9fc3c8605e58a7399a70127b39b1.png)
, such that 
Prove that :
equality cases : ?
Y by nguyendangkhoa17112003, Adventure10, Mango247, Mango247, Mango247, and 1 other user
The problem's inspired from the other, but this one is unsolved now. It looks very simple, but really difficult!
Prove that for all
are non-negative reals we have the inequality :
I hope I 'll see a nice solution for it!
Good luck every one!
To Vasc : I think You 've thought about it? Do you have a solution?
Prove that for all
are non-negative reals we have the inequality :
I hope I 'll see a nice solution for it!
Good luck every one!
To Vasc : I think You 've thought about it? Do you have a solution?
is the same as to prove one of this two 
and/or 
I could be wrong 
and use the inequality 
.
with 
using PM-AM 
I'll have to think more to get the stricter inequality...
,
.
,
.
are nonnegative reals we have the inequality :
![$ \left[11x^3 + 28x^2(y + z) + 7x\left(y^2 + 6yz + z^2\right) + 2(y + z)\left(y^2 + 10yz + z^2\right)\right]^2$](http://latex.artofproblemsolving.com/0/e/2/0e241989b33557ca49488f3abeeb541003beb3a3.png)
![$ - 4\left(x^2 + yz\right)\left[5\left(x^2 + y^2 + z^2\right) + 14(yz + zx + xy)\right]^2$](http://latex.artofproblemsolving.com/3/6/5/365f58e072a95c778f0c2a845a8b13b6db324018.png)
![$ = \frac {\left[63x^3 + 84x^2(y + z) - x\left(125y^2 - 284yz + 125z^2\right) - 22(y + z)(y - z)^2\right]^2}{189}$](http://latex.artofproblemsolving.com/b/8/e/b8e86e034d3838c411fd59193c9fc088bf7adcbc.png)
![$ + \frac {4(y - z)^2\left[x^2(y + z)\left(68y^2 + 15yz + 68z^2\right) - 2x(y - z)^2\left(68y^2 + 117yz + 68z^2\right) + (y + z)(y - z)^2\left(68y^2 - 53yz + 68z^2\right)\right]}{189(y + z)}$](http://latex.artofproblemsolving.com/9/d/a/9da46bfbc9bee644b4022331c40efcccddd5f18c.png)
.
,
.
,
.
,![\[\frac{3}{2}\,\sqrt { \left( x+y+z \right) ^{2}-9\,{\frac { \left( xy+zx+yz
\right) xyz}{ \left( x+y+z \right) ^{3}}}}\geq \sqrt {{x}^{2}+yz}+\sqrt {
{y}^{2}+zx}+\sqrt {{z}^{2}+xy}\]](http://latex.artofproblemsolving.com/7/0/2/702cf64c0520cb076d8fba7768a6461bfb7eb736.png)
Use (AM-GM) we have :
