AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
classes and 
student. We know that in this school every two students have attended exactly in one common class. Also due to smallness of school each class has less than students. If 
is not a perfect square, prove that there exist a student that has attended in at least 
classes.
, prove that there exists a positive integer 
, such that for any integer 
, can be expressed by a sum of positive integers 
's as![\[n=a_1+a_2+\dots+a_m,\]](http://latex.artofproblemsolving.com/6/c/2/6c2a979a5ff6c40bf2d7152c8c32088fc36f848b.png)
where 
, 
, 
, 
and 
.
where is an odd integer. Let 
be a function defined on 
taking values in 
such that
for all 
for all 
; b) 
identical-looking coins in a circle. It is known that exactly two of them are fake. Real coins weigh the same, fake ones too, but they are lighter than the real ones. How can you determine in three weighings on a cup scale without weights whether there are fake coins lying nearby or not??
points of general position marked on the coordinate plane (i.e., points among which there are no three lying on the same straight line). Is there a polynomial of two variables 
a) of degree 
; b) of degree such that it equals to zero exactly at these marked points?
be triangle, inscribed in parabola. Tangents in points 
forms triangle 
. Prove that 
. (
is area of triangle 
).
such that there exist integers 
with ![\[x^3_1+x^3_2+\,\ldots\,+x^3_t=2002^{2002}\,?\]](http://latex.artofproblemsolving.com/e/f/a/efaa98e77ac0273284ee3446e4335d8cc4f681cb.png)
for 
for positive reals 
.
for positve reals .
![\[ a^4 + b^4 = 4ab - 2\]](http://latex.artofproblemsolving.com/1/5/6/156f7be5ac3d886491d230194b9e2c6400797ef3.png)
in positive reals.
for for positive reals .
for for positive reals .
when 
for for positive reals . for positve reals .
for 
positive integers.
are positive reals with 
. Prove![\[ \sum_{cyc} \frac{1}{a^3(b+c)} \ge \frac{3}2\]](http://latex.artofproblemsolving.com/3/2/7/3274c741499c72d0164802a7864d8e5577fc81b4.png)
are non negative reals such that 
. Prove![\[ a^3+b^3+c^3+6abc \ge \frac{1} 4 \]](http://latex.artofproblemsolving.com/a/9/a/a9a48560ffaa2cb72ed3688af8402e0b56e52a69.png)
for for positive reals . for positve reals .
and 
.![\[P=\frac{1}{5-2a}+\frac{1}{5-2b}+\frac{1}{5-2c}.\]](http://latex.artofproblemsolving.com/c/3/7/c3708768a84c2bbb1e4fae502a9bb1e92cb13d02.png)
and 
Prove that
and 
Prove that
.
and 
.
.
,
.
.
,
and 
.
.
cancels out all terms, 
.
.
which is true from applying 6-variable AM-GM.
,
because rearranging gives 
which is true from trivial inequality. So we want them all to be equal, which is only possible when 
(because of AM-GM"s equality condition. Then we must have 
.
and 
.![$[(x+y)(y+z)(z+x)]^2\ge [8xyz]^2\Rightarrow (x+y)(y+z)(z+x)\ge 8xyz$](http://latex.artofproblemsolving.com/4/3/6/43647338b5e7ba424736e600f0f592181560bfe0.png)
by taking positive square roots.
for positive a,b,c
or 
when ![\[\left(\dfrac{a^3+b^3+1}{ab}\right)\left(\dfrac{b^3+c^3+1}{bc}\right)\left(\dfrac{c^3+a^3+1}{ca}\right).\]](http://latex.artofproblemsolving.com/7/4/6/7463c2e4cfc9d3e31740a16a27e958e5c3d55cec.png)
By AM-GM, we have ![$\dfrac{a^3+b^3+1}{ab}\geq\dfrac{3\sqrt[3]{a^3b^3}}{ab}=3$](http://latex.artofproblemsolving.com/1/7/5/175dd3c56948cef8fd9fa1d05b8c20262f2ac9ee.png)
.
.
,![\[ x + y \ge 2\sqrt{xy} \\ y+z \ge 2\sqrt{yz} \\ z+x \ge 2\sqrt{zx} \]](http://latex.artofproblemsolving.com/e/3/8/e380b0a520f67702472eb8da66a079672870accb.png)
Multiplying the three inequalities gives 
from four-variable AM-GM. Therefore 
.
or 
,
.
where 
is a positive real number.
by five variable AM-GM.![\[\dfrac{x+x+x+x+\dfrac{1}{x^4}}{5} \ge \sqrt[5]{x \cdot x \cdot x \cdot x \cdot \dfrac{1}{x^4}} = 1\]](http://latex.artofproblemsolving.com/c/4/b/c4bbdc916ad6dddf472840539bb11d262d89ea10.png)
so 
.
.
gives ![\[\sum_{cyc}\frac{1}{a^3(b+c)}=\sum_{cyc}\frac{x^2}{y+z}.\]](http://latex.artofproblemsolving.com/5/d/8/5d826cf943840687ca0b641ce461693f0a8a1bdb.png)
From AM-GM we have 
and ![$x+y+z\geq 3\sqrt[3]{abc}=3$](http://latex.artofproblemsolving.com/3/7/5/375d6e0603a4db834ee98a734695342c534eb696.png)
,![\[\sum_{cyc}\frac{x^2}{y+z}\geq \sum_{cyc}\left(x-\frac{y+z}{4}\right)=\frac{x+y+z}{2}\geq \frac{3}{2},\]](http://latex.artofproblemsolving.com/f/3/4/f34ff0447752d3366a262073882e04fe1bd5e38c.png)
as desired. From the nature of AM-GM, equality holds at 
.
which is weaker than and immediately follows from Schur. For equality, from Schur we need 
,
are 
and the other is 
.