AoPS Academy
be positive reals with 
. Then show that![\[\sum_{i=1}^n \frac{x_i}{2-x_i} \ge \frac{n}{2n-1}\]](http://latex.artofproblemsolving.com/f/6/5/f65f6fce4465125dfacf002f0659c898df7ae150.png)
Prove that: 
is the sum of 
square numbers
Let 
and 
is the sum of square numbers. Prove that: 
is the sum of square numbers
+1 w
and 
are the incenter and the 
-excenter of triangle 
, respectively.
be the midpoint of arc 
on the circumcircle, and 
be the foot of the to 
. The line 
intersects the circumcircle again at 
. For
on the circumcircle of triangle , let 
intersect at 
. Prove
are concyclic.
Also great title.
is odd. 
is even and 
is not even. Thus is even. If 
, its clear that 
, by second and thirth condition 
, contradiction. So 
. If 
so 
. And by second condition 
is a divisor too. If 
by firs t condition we get 
and 
, if 
is not a divisor and 
is, we get 
and , if 
and doesnt, 
and 
but is not a divisor. if and both are not divisors, we get 
and 
and we know that 
is a prime thus it must be 
. 
. If 
Let 
If 
. By Last two conditions we get 
gives a contradiction. Thus 
so 
are primes. And by is even its 
thus we get 
and ![$\Delta=[2p(p-4)]^2-4\cdot 4p^2+4$](http://latex.artofproblemsolving.com/c/e/2/ce28fa3d1cff5202692b30ff64666425d190e3e2.png)
and it can't be perfect square eqsily. Thus theres no solution more. Thus 
.
.
works. The divisors of are 
which indeed satisfy
is a multiple of
is odd, then all of its divisors will be odd, so 
, a contradiction. Thus, is even. Since 
is an odd perfect square, it must satisfy 
, the claim follows. A corollary is that 
.
and .
![\[
(d_6-d_7)^2=n+1-\frac{4n}{d_3}
\]](http://latex.artofproblemsolving.com/d/4/b/d4b5c12a7adf13da6f8271b47e5649f684038353.png)
If 
then the right hand side would be strictly smaller than 
, a contradiction. Thus 
. As 
we must have that . Thus 
and hence . A consequence of the fact that is that 
.
and into the first equation yields![\[
3 + d_4 + d_5 = d_6.
\]](http://latex.artofproblemsolving.com/2/0/7/207a22d94b04115b009f326e57abe5d8d10afa43.png)
We proceed with casework on 
.
(and thus a prime smaller than )
, skipping .
. But 
.
, but .
is even
, but then 
, a multiple of
is odd
. As is the first odd divisor, it must be prime so either 
or 
. The only possibility is 
giving 
. Then 
, as desired.
. We start with the following observation. and . were odd then 
and are all odd. But then,![\[d_7 = 2d_2+d_4+d_5 \equiv 0 + 1 + 1 \equiv 0 \pmod{2}\]](http://latex.artofproblemsolving.com/b/e/a/bea61628b07c20ceed6747b2e236443a7a26c702.png)
which is a contradiction. Thus, is even which implies . Further, since is even,![\[(d_6+d_7)^2 = n+1 \equiv 1 \pmod{2}\]](http://latex.artofproblemsolving.com/2/a/a/2aa0ae016da690d73ede18945bd8b7bcc71fcba2.png)
so 
is even. Since it is well known that the square of an odd integer is always 
it follows that is divisible by . Thus, 
so or . Note that if ,
But by AM-GM we have, 
which is a clear contradiction. Thus, which proves the claim. holds for any satisfying the given conditions., letting 
we have
which implies as claimed.
and the square of an integer is always 
or 
, since the only possibility is 
so 
which implies 
.
, is a factor of . Further since , 
. Thus, is one of 
, , 
and . These will be our four cases.
. We then require 
, 
, 
and , which is a contradiction since .
. But this is a contradiction since 
.
. Then we require, or 
(since 
). If , then 
so . This immediately gives 
which is a contradiction since this is not divisible by . If , then we have 
and which is a contradiction since .. Here we have two possibilities. If , then 
and which is divisible by 3 which is impossible. If there exists some odd prime 
, 
so 
. Since 
. So 
or 
., 
which is impossible since .
, 
so 
. Since no multiple of proceeds we must have . So and 
. Then, 
. It now remains to write out the divisors of and check that all the conditions are met, which is easy enough.
is even it's contradiction. I mean 
and .
and 
,we can see the only solution is
and it is contradiction.
and 
. So 
.İt means 
.
but 
.İt is contradiction. we have 2 option. Firstly 
but 
is contradiction. Secondly 
but is contradiction. I mean 
. So 
. If 
but is contradiction. I mean 
. The only solution is .
denote all positive divisors of 
,
even, which means 
,
.
is odd, and 
is odd, which means 
.
,
,
,
,
.
.
is 
.
,
fails by checking. If 
,
fails by checking.
,
,
.
,
,
,
,
,
,
.
,
,
,
,
,
,
.
,
s. Then, 
is odd and a perfect square, it must be in the form 
hence 
.
,
.
is 
but 
,
,
so 
but 
,
so 
but 
also divides 
but 
.
where 
is a prime number greater than 
is an odd number so 
is even. Then, 
.
so 
.
is odd. But 
.
.
then 
and 
then 
so 
that gives us 
and 
but 
.
when 
that means 
but 
when 
and 
we can get 
and 
.
and 
then 
and 
.
.
.
.
then 
when 
then 
and 
.
therefore
That means 
