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.
such that, for all positive integers 
and 
,
be a positive real number. Find all functions 
such that 
and![\[
f(af(x) - f(y) + y) \leq x + f(y) - y, \quad \forall x, y \in \mathbb{R}^+.
\]](http://latex.artofproblemsolving.com/e/2/9/e296b71ad8b8bd977e2bb2a3630a5bc020b8e03a.png)
, let 
be the least prime that does not divide 
, and define 
to be the product of all primes less than . In particular, 
For having 
, define 
. Consider the sequence 
defined by 
and ![\[ x_{n+1} = \frac{x_n p(x_n)}{q(x_n)} \]](http://latex.artofproblemsolving.com/8/2/5/825be3bbef6a29de69aead03392ca7215093f320.png)
for 
. Find all such that 
.
be a natural number. At a mathematical olympiad training camp the same 
courses are organised every day. Each student takes exactly one of the courses each day. At the end of the camp, every student has takes each course exactly once, and any two students took the same course on at least one day, but took different courses on at least one other day. What is, in terms of , the largest possible number of students at the camp?
from the set of positive integers to the set of positive integers such that, for all positive integers 
and 
, there exists a non-degenerate triangle with sides of lengths![\[ a, f(b) \text{ and } f(b + f(a) - 1).\]](http://latex.artofproblemsolving.com/4/a/b/4ab97da27e05d0f9e7762f74d0d0bd2dc8b9842f.png)
and 
be the incircle and circumcircle of the acute triangle 
, respectively. Draw a square 
so that all of its sides are tangent to , and 
, 
are both on 
. Extend 
and 
, intersecting at 
and 
, respectively. Prove that 
and 
intersects on .
, then prove that the following inequality holds:![\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]](http://latex.artofproblemsolving.com/9/5/5/955863a0cf8747ae45b736b0631f243d3908eb84.png)
larger than 
, solve the system of equations (assuming 
, for 
):![\[ \begin{cases} \displaystyle x_1+ \phantom{2^2} x_2+ \phantom{3^2} x_3 + \cdots + \phantom{n^2} x_n &= n+2, \\ x_1 + 2\phantom{^2}x_2 + 3\phantom{^2}x_3 + \cdots + n\phantom{^2}x_n &= 2n+2, \\ x_1 + 2^2x_2 + 3^2 x_3 + \cdots + n^2x_n &= n^2 + n +4, \\ x_1+ 2^3x_2 + 3^3x_3+ \cdots + n^3x_n &= n^3 + n + 8. \end{cases} \]](http://latex.artofproblemsolving.com/0/8/f/08fc55e0cc886c4c44f678e75eb07e4d814c4a77.png)
be a triangle. On the sides , 
, 
of the triangle, construct outwardly three squares with centres 
, 
, 
respectively. Let be the circumcircle of 
. Given that 
lies on , prove that the centre of lies on the perimeter of 
.
be an integer, and let 
and 
be sequences of non-negative integers. For each integer 
, let 
and 
, where 
is the integer such that 
is divisible by . We say is -harmonic if each 
equals the following arithmetic mean: ![\[a_i = \frac{1}{2b_i+1} \sum_{s=-b_i}^{b_i} a_{i+s}.\]](http://latex.artofproblemsolving.com/1/1/7/117ca4963c8e443fd3cbfb55a907bee4be295151.png)
Suppose that neither 
nor is a constant sequence, and that both is -harmonic and is -harmonic.
of the numbers 
are zero.
is given on the plane. Let the points 
, 
and 
be on the sides , and , respectively. (, and are different from the vertices , 
and 
) Also the points , and 
are taken such that 
is an equilateral hexagon. Suppose that the circumcenters of and 
are coincident. Then determine the least possible value of:
Note: 
area of 
denote the sides of a triangle and ![$[ABC]$](http://latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
the area of the triangle as usual.
If ![$6[ABC] = 2a^2+bc$](http://latex.artofproblemsolving.com/e/2/8/e28bc8b32d1e9af224bc600821a93ba8caf02d08.png)
, determine 
.
For all triangles, prove that ![$3a^2+3b^2 - c^2 \ge 4 \sqrt{3} [ABC]$](http://latex.artofproblemsolving.com/6/a/e/6ae3a2a3a6082de0672e1da75f9a84a3cdd342ba.png)
.
. We fix and the length of the altitude from , then the area is fixed. Note that 
is minimum when 
, so the ![$6[ABC]=3a\sqrt{b^2-\frac{a^2}{4}}$](http://latex.artofproblemsolving.com/1/7/9/1796f2c0880833af957124175c8a7388a38ab62d.png)
. Then![\[3a\sqrt{b^2-\frac{a^2}{4}}\leq 2a^2+b^2\iff 9a^2b^2-\frac94a^4\leq 4a^4+b^4+4a^2b^2\iff 5a^2b^2\leq \frac{25}{4}a^4+b^4
\]](http://latex.artofproblemsolving.com/5/8/6/586aba08ae34090d9c8189cd2ae907ac64e562ee.png)
which is true by AM-GM with equality iff 
, or 
.
and 
Prove that
.
