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.
and 
Prove that
Where 
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?
be an integer, and let 
be positive real numbers such that 
. Prove that![\[(1 + a_2)^2 (1 + a_3)^3 \dotsm (1 + a_n)^n > n^n.\]](http://latex.artofproblemsolving.com/7/7/f/77fb719e975225b59c64f1f3aa102c98a6c25f50.png)
not be a trapezoid such that there is a circle centered at that is tangent to the four sides 
, 
, 
, and 
. Let , , 
, and 
be the circumcenters of the triangles 
, 
, 
, and 
, respectively. Prove that there is a circle containing the circumcenters of the triangles 
, 
, 
, and 
.
such that
satisfies 
. Find the value of 
.
be positive real numbers so that 
. Prove that![\[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1.
\]](http://latex.artofproblemsolving.com/7/6/e/76e794a62090c8c4b5f3a70ad4d036669418ca69.png)
, there exist positive real numbers 
, 
and 
such that 
, 
and 
. The inequality then reduces to![\[ \left( \frac{x-y+z}y \right) \left( \frac{y-z+x}z \right) \left( \frac{z-x+y}x \right) \leq 1. \]](http://latex.artofproblemsolving.com/d/a/6/da6f9bad78d4d92814b774a070ec5a199546e61a.png)
, 
and 
. The inequality then reduces to ![\[ 8pqr \le (p+q)(q+r)(r+p) \]](http://latex.artofproblemsolving.com/a/0/e/a0e912dc9dde61f92dd1e0d75fcdd321ecf6e623.png)
which is obvious from the AM-GM since at most one of 
can be negaitve..
so we have to prove this
then 
and 
an others then it ia sides of triangle so take
so from it follows. (we will look next step)
. Since 
are positive so
![\[8pqr \le (p+q)(q+r)(r+p)\]](http://latex.artofproblemsolving.com/e/3/9/e39bf0011b2a9ce9a212ac340670d747d32e60fb.png)
which is obvious from the AM-GM since at most one of can be negaitve..
are positive by definition. can be negative. If one of the numbers is negative, then it is trivial since the left side would be negative and the right side positive.
for example? If not, you can't appy AM-GM. There is a solution in that sense, but not as trivial as the one posted there.
Where 
such that 
.
and multiply each factor
.![\[ \left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leq 1\]](http://latex.artofproblemsolving.com/5/2/a/52a6b239248cfe5040e44a667dfc5987c4d5a1bb.png)
(Ji chen)
,
,
.![\[\prod_{\text{cyc}} \frac{x+z-y}{y} \le 1\]](http://latex.artofproblemsolving.com/0/9/f/09f3e716c14f66f17353de5ce3d6384ee4b1e749.png)
Expanding, it suffices to prove that![\[\sum_{\text{cyc}} x^3 + 3xyz \ge \sum_{\text{sym}} x^2y\]](http://latex.artofproblemsolving.com/5/b/8/5b8c481426bfdb576e06cb4341e589c78715b91d.png)
which is true by Schur.
)
and cyclic variants, then we can rewrite as 
.
and cyclic variants so 
are sides of a triangle, set 
and cyclic variants, then we are reduced to proving 
,
with 
.![\[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1 \iff (x+y-z)(x-y+z)(-x+y+z) \le xyz.
\]](http://latex.artofproblemsolving.com/d/b/9/db9b2e605e5c3b3250ee9ff79f947e076c2af769.png)
Now we have two cases:
which are negative is at least 
.
then adding yields a contradiction.
Case 3: All of the terms 
.
.
which is true by AM-GM.
with 
.
Now by using the substitution 
we can homogenize the above inequality. Thus we have to show that,
which is trivial by Schur's and hence, we are done. 
and so on. Multiply the product of denominators on both sides of the inequality and expand the 
and simplify a bit. The resulting inequality is just a case of Schur's.
We wish to show that 
or 
Expanding gives Schur's inequality, which is true.
such that:![\[
a = \frac{x}{y}, \quad b = \frac{y}{z}, \quad c = \frac{z}{x}.
\]](http://latex.artofproblemsolving.com/4/6/4/464af3e2af4fb20cab9fae3d17ad0ce7dfadd50c.png)
Thus, the given inequality transforms into:![\[
(x + y - z)(-x + y + z)(x - y + z) \leq xyz.
\]](http://latex.artofproblemsolving.com/9/b/9/9b9cb9788281fec4f7c4b6dd1a5c81cf2701ff12.png)
Expanding this, we obtain:![\[
x^3 + y^3 + z^3 + 3xyz \geq x^2(y + z) + y^2(x + z) + z^2(x + y).
\]](http://latex.artofproblemsolving.com/b/7/3/b73dc791b148ac9993bad4f8ea7f4e5250c6ba15.png)
which is trivial by Schur's inequality
is almost essential to the problem. This basically simplifies out to 
.
,
,
and 
WLOG then we have 
which fails since x,y,z all must be positive. But if 
then the left hand side must be non-positive, and hence the negatives case is considered. We move on by assuming the Left Hand Side is positive.
,
,
.
through ravi substitutions (yay). This becomes 
and therefore we take AM-GM on individual factors in the RHS which finishes the problem.