and 
be two sequences defined by:![\[
a_{n+1} = \frac{1 + a_n + a_n b_n}{b_n} \quad \text{and} \quad b_{n+1} = \frac{1 + b_n + a_n b_n}{a_n}
\]](http://latex.artofproblemsolving.com/a/a/0/aa0a5c5acedc990eb345fb50a8395db05307a4e7.png)
for all 
,
and 
.![\[
a_{2024} < 5.
\]](http://latex.artofproblemsolving.com/f/c/a/fca6e6630f33429b5a7ea867d639a2e29ff881f2.png)
children, no two of the same height, lined up. Let us call a pair of different children 
good if between them there is no child whose height is greater than the height of one of 
and 
,
are considered the same pair.)
satisfy 
prove in equality.
![\[\sum_{cyc}(x+y)\sqrt{(z+x)(z+y)} \geq 4(xy+yz+zx),\]](http://latex.artofproblemsolving.com/0/4/7/047adb616a6be9f9ed52f2385fa592fb5464e420.png)
and 
.
:D :play_ball:
such that![\[\lim_{n \to \infty} \left( \frac{1 + 2^{1/3} + 3^{1/3} + \dots + n^{1/3}}{n^t} \right ) = c\]](http://latex.artofproblemsolving.com/e/1/a/e1a87bd7bee41a7d656e2ff067a87dff0c9929ec.png)
for some real 
.
.
![$f\in C^1[a,b]$](http://latex.artofproblemsolving.com/8/4/8/848da12534d564560fe1cb016142159234be98b4.png)
,
and suppose that 
,
,
for all ![$x\in [a,b]$](http://latex.artofproblemsolving.com/7/8/d/78d9dcd969a2c45f7cc9234aa1377b6ec1c2d5d3.png)
.
for all 
is compact?
Y by Ankoganit, anantmudgal09, ATimo, rafayaashary1, ahmedAbd, laegolas, kk108, Tawan, Adventure10, Mango247, sami1618
In the country of Sugarland, there are 
students in the IMO team selection camp.
team selection tests were taken and the results have came out. Assume that no students have the same score on the same test.To select the IMO team, the national committee of math Olympiad have decided to choose a permutation of these tests and starting from the first test, the person with the highest score between the remaining students will become a member of the team.The committee is having a session to choose the permutation.
Is it possible that all students have a chance of being a team member?
Proposed by Morteza Saghafian
students in the IMO team selection camp.
team selection tests were taken and the results have came out. Assume that no students have the same score on the same test.To select the IMO team, the national committee of math Olympiad have decided to choose a permutation of these tests and starting from the first test, the person with the highest score between the remaining students will become a member of the team.The committee is having a session to choose the permutation.Is it possible that all
students have a chance of being a team member?Proposed by Morteza Saghafian
This post has been edited 3 times. Last edited by bgn, Apr 7, 2017, 6:42 PM
il all the tests. He has no chance for being a team member?
,
students. But don't have an idea about the maximum number.
can get into the team with the order 
.
This is in fact possible, with the following setup, where each column represents the top 6 scorers on each problem, and the contestants are numbered 
.
t
t
:
into the team first, you have to get 
from Test 
from Test 
before you start taking 
,
corresponds to the rank of student 
on problem 
can get on the team as if the problem they are ranked first on is chosen first;
;
;
can get on the team if the problems are chosen in the order 
;
can get on the team by symmetry (with 
replaced with 
and so on).
;
can get on the team if the problems are chosen in the order 
;
.
square to easily be able to eliminate top ranks; then, for each pair, we would be able to uplift three students who were ranked lower, which suffices for the 
satisfies
through 
.
would be admitted as 
is admitted.
This is not necessary but it is sufficient in that every student that appears at the bottom of a pyramid (size 
It is not difficult to verify how this works.
the first row (rank) on which the 
appeared, this was done as I saw that if a person is on the 5th rank then for that guy to be picked up by there we must have a sort of upside-down pyramid, like
Here, if 
is to be picked from the bottom row, then we must have that pyramid shape above it (not necessarily in order, but you get it).
over all men(essentially extremal principle) and case bashed on its values (i.e., when it is 2, 3,... 6). Case of 2 was trivially false, but I tried 3 and saw that I could just put 3 people twice in row 2, and then if person 1 fills row 1 completely, then we can get up to 10 people. ( the reason to do this was that I wanted to put people in row three, i.e as 
![\[\begin{array}{|c|c|c|c|c|c|}
\hline
1 & 2 & 3 & 4 & 5 & 6 \\
\hline
A & A & A & A & A & A \\
\hline
B & B & B & B & B & B \\
\hline
C & C & C & C & C & C \\
\hline
D & D & D & D & D & D \\
\hline
E & E & F & F & G & G \\
\hline
H & I & J & K & L & M \\
\hline
\end{array}\]](http://latex.artofproblemsolving.com/c/8/2/c822faec25af888c2bf8e08367c205ee86117993.png)
