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.
be the circumcircle of acute triangle 
. Points 
and 
are on segments 
and 
respectively such that 
. The perpendicular bisectors of 
and 
intersect minor arcs and of at points 
and 
respectively. Prove that lines 
and 
are either parallel or they are the same line.
of them there were either 
who were all friends among each other or who weren't friend with each other. Prove that you can separate all the people at the party in two groups in such a way that in the first group everyone is friends with each other and that all the people in the second group are not friends to anyone else in second group. (Friendship is a mutual relation).
and 
play a game, given an integer 
, writes down 
first, then every player sees the last number written and if it is 
then in his turn he writes 
or 
, but his number cannot be bigger than . The player who writes wins. For which values of does win?
be the circumcircle of a scalene triangle . Let 
be a circle internally tangent to in . Tangents from touch in 
and 
, such that lies in the interior of 
. Similarly, tangents from 
touch in 
and 
, such that lies in the interior of .
and 
concur on the bisector of 
.
, let 
be all positive integers smaller than that are coprime to . Find all 
such that 
for all 
is the largest positive integer that divides both 
and 
. Integers and are coprime if 
., 
and 
are the midpoints of sides 
and respectively. Point lies inside the triangle. Let 
. Prove that 
.
, 
, and 
of a hexagon 
inscribed in a circle 
intersect at a point 
, and the acute angle between any two of them is 
. Let 
be the radius of the circle tangent to segments 
and 
and internally tangent to ; the radii 
, 
, 
, 
, and 
are defined similarly. Prove that![\[
r_{AB}r_{CD} + r_{CD}r_{EF} + r_{EF}r_{AB} = r_{BC}r_{DE} + r_{DE}r_{FA} + r_{FA}r_{BC}.
\]](http://latex.artofproblemsolving.com/0/d/6/0d66d6685bbc97c7553e9ed3ba14ea0dc3328b49.png)
, let 
be the circumcentre, 
be the orthocentre, and be the centroid.
satisfy the following condition:
is constant.
. Find .
such that numbers 
are both perfect squares.
such that ![\[\left(g(m)+n\right)\left(g(n)+m\right)\]](http://latex.artofproblemsolving.com/5/a/c/5ac9fce315e8a524a3a01b345cfe76bc76056d68.png)
is a perfect square for all 
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.
and 
Prove that
Let 
Prove that
and 
.![$\sqrt[3]{ab-1}+\sqrt[3]{bc-1}+\sqrt[3]{ca-1} \le \sqrt[3]{(a+b+c)^2}$](http://latex.artofproblemsolving.com/9/6/3/9633b5849794285e44519c745c1f616eceb77711.png)
and multiply each factor
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.
hours
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.
,
,
.
are side lengths of a triangle. Write 
through ravi substitutions (yay). This becomes 
and therefore we take AM-GM on individual factors in the RHS which finishes the problem. 
for positive reals ![\[
(x - y + z)(y - z + x)(z - x + y) \le xyz.
\]](http://latex.artofproblemsolving.com/7/6/4/76461f23eb9524e2107325ece81d4ba2d8e26750.png)
Now, notice that if 
,
and 
for positive reals 
.![\[
8def \le (d+e)(e+f)(d+f).
\]](http://latex.artofproblemsolving.com/a/4/e/a4e0db3a479faef7aca7659bca20159cce238b7f.png)
Finally, by AM-GM, we have that 
,
and 
.
,
.![\[\frac{(x+y-z)(-x+y+z)(x-y+z)}{xyz} \le 1 \iff \sum_{\text{sym}} x^2y \le x^3+y^3+z^3+3xyz.\]](http://latex.artofproblemsolving.com/0/f/a/0fa9bb71f48dc8bdd60444915d3dcc99ef8f186a.png)
But this is Schurs with 
.
Hence, we have 
By expanding, we have that 
This follows from Schur's.
.
which is Schur's inequality for 
such that
Substituting into the original problem statement, we have to prove that
or equivalently, that
where we derive the second inequality by multiplying the first by 
.
,
,
.
(and all other cyclic variations) and we must prove that
which immediately follows from performing AM-GM three times over the pairs of any two of 
.