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 a fixed positive integer. There are 
people, numbered 
, participating in a tennis tournament. For any two positive integers 
with 
, player 
has a higher skill level than player 
. Prior to the first round, the players are paired arbitrarily and each pair is assigned a unique court among 
courts, numbered 
.
, the winner of court moves to court 
and the loser of court stays on court ; however, the winner of court 1 stays on court 1 and the loser of court 1 moves to court .
such that, regardless of the initial pairing, the players 
all change courts immediately after the th round.
+2 w
be positive reals. Prove that![\[ \frac{(a-b)(a-c)}{2a^2 + (b+c)^2} + \frac{(b-c)(b-a)}{2b^2 + (c+a)^2} + \frac{(c-a)(c-b)}{2c^2 + (a+b)^2} \geq 0. \]](http://latex.artofproblemsolving.com/6/9/3/6937791380a8642edca3e773dc72dbf3f0e0f206.png)
be a perfect square then, is it true that it must be of the form
be points on the sides 
respectively of a triangle 
. Suppose that 
and 
meet at 
and meet at 
, and 
and meet at 
, such that 
, and 
. Compute the ratio of the area of 
to the area of 
.
and 
intersect at different points 
and 
. The straight lines tangents to that pass through and intersect at 
. Let be a point on that is out of . The 
line intersects at 
again, the 
line intersects again to in 
and the line 
intersects again to the circumference in 
. Prove that the 
line passes through the midpoint of the 
segment.
are positive real numbers such that 
, prove that ![\[ (1+a)(1+b)(1+c) \geq 8 (1-a)(1-b)(1-c) . \]](http://latex.artofproblemsolving.com/0/b/e/0bedd9ed8362ac5202845f8a4d0d24519423e9b4.png)
with 
, points 
are such that 
and 
and 
, and 
and 
, as shown on the picture. Suppose that 
is a right angle. Prove that the points 
are collinear.
. Prove that![\[
\sum_{\text{cyc}} \frac{x^3}{y^2 + z^2} \geq \frac{x + y + z}{2}
\]](http://latex.artofproblemsolving.com/9/0/3/903220c471bdd3028dda2fcb13874a4bcf752872.png)
without using the Rearrangement Inequality or Chebyshev's Inequality. be an acute angled triangle.And altitudes 
and 
intersects at point 
.Let 
be a point on ray such that 
.Circumcircle of 
intersects line 
at and so prove that 
,![\[\frac{1}{a+\frac{1}{b}+1}+\frac{1}{b+\frac{1}{c}+1}+\frac{1}{c+\frac{1}{a}+1}\ge \frac{3}{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1}. \]](http://latex.artofproblemsolving.com/4/3/1/431b0ed375e037ee3d19e48a2bb4099f12f87784.png)
Proposed by David Stoner
such that![\[ x f(x + xy) = x f(x) + f \left( x^2 \right) f(y) \quad \forall x,y \in \mathbb{R}.\]](http://latex.artofproblemsolving.com/b/0/f/b0f1a1d3b9de5d23f1b5a2e9be3b88d8bd575883.png)
is not divisible by 
for any positive integer 
.
, s.t :
, there exists some positive integer 
that makes the following equation have no integer root 
.
are nonnegative because 
. We aim to show that there must exist a less than 
that cannot be equal to the left-hand side of the given equation. We count the number of possible values of each so that 
., assume for the sake of contradiction that 
. When we multiply this by itself 
times we see that 
, which we don't want. to be less than 
. There are possible values for in the range 
. When we take the product of for all such that 
, we get ![\[ \prod\limits_{i=1}^n 2^{2^{n-i}} = 2^{\sum\limits_{i=1}^n 2^{n-i}} = 2^{2^n - 1}\]](http://latex.artofproblemsolving.com/e/4/d/e4d0e5a2f1977c4e12bec6fa560fbe9166f30418.png)
This value is an upper bound for the amount of tuples 
such that the LHS of the given equation is less than . However, this value 
is less than the 
possible values of less than . Thus, by pigeonhole there must exist a less than that cannot be equal to the LHS of the given equation.
,such that 
P
is always false. If it turns out this equation can be true for some large