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.
denote the set of positive integers
such that for all 
it holds that 
Note: consider 
and 
and ![$P_1(x),P_2(x), \ldots P_n(x) \in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/c/6/4/c64176e52d94032e571b4579e35f9e6adfde675a.png)
such that 
. Prove that exists many 
such that every equation: 
has no real roots
with 
. Points 
and 
are on 
and 
respectively such that 
. Lines 
and 
meet at 
. 
is the incenter of 
and 
is the orthocenter of 
. 
and 
are circles with diameter 
and 
, respectively, intersecting each other at points 
and 
. Prove that and lie on 
.
. Prajit then tries to place n points such that no four points are concyclic and no 
points are collinear in Euclidean plane, such that Shining cannot find a group of three points such that their circumcircle contains none of the other remaining points. Is he always able to do so?
. Otherwise the given inequality is false.
and expand.
by AM-GM. Hence it suffices to show that 
.
.
, and 
.
.
,
The four equality cases occur when 
or when two of 
are equal and the third is 
. and (c) is combining (b) with 
which comes from AM-GM.
do algebra with just the 1st two terms. What about the third term? What if the assumption is not true?
was given to be positive in the paper, rather than non-negative)
such that
for all 
.
and 
Prove that
Where 
Where 
such that 
.
prove that![\[a^r(a-b)(a-c) + b^r(b-a)(b-c) + c^r (c-a)(c-b) \geq 0 \]](http://latex.artofproblemsolving.com/0/6/c/06cdb779fc271d73a80bdec409c777e833dd2980.png)
(b) Find an inequality for non negative 
on the greater side.
for non negative 
as 
and 
Therefore the sum of the first two terms is positive. The third term is obviously positive. We are done.
Use 
,
(homogenize it)