be distinct real numbers such that![\[
a + b + c + \frac{1}{abc} = \frac{19}{2}
\]](http://latex.artofproblemsolving.com/5/1/e/51eef77358db2b6c98f0fb0fb4e30db837a03642.png)
Find the maximum possible value of 
.
such that for all positive real numbers 
and 
,![\[
\frac{x + f(y)}{x f(y)} = f\left( \frac{1}{y} + f\left( \frac{1}{x} \right) \right)
\]](http://latex.artofproblemsolving.com/7/d/f/7df25755bc33bef970fffb392fb5b6b789390881.png)
.The solution to 
is only 
for some 
Y by Functional_equation, centslordm, InternetPerson10, Mathmick51, TheStrayCat, megarnie, Aritra12, 554183, adityaguharoy, Aopamy, yshk, deplasmanyollari, eggymath, Kingsbane2139, Rounak_iitr, Funcshun840, cubres
Two squirrels, Bushy and Jumpy, have collected 2021 walnuts for the winter. Jumpy numbers the walnuts from 1 through 2021, and digs 2021 little holes in a circular pattern in the ground around their favourite tree. The next morning Jumpy notices that Bushy had placed one walnut into each hole, but had paid no attention to the numbering. Unhappy, Jumpy decides to reorder the walnuts by performing a sequence of 2021 moves. In the 
-th move, Jumpy swaps the positions of the two walnuts adjacent to walnut .
Prove that there exists a value of such that, on the -th move, Jumpy swaps some walnuts 
and 
such that 
.
-th move, Jumpy swaps the positions of the two walnuts adjacent to walnut .Prove that there exists a value of
such that, on the -th move, Jumpy swaps some walnuts 
and 
such that 
.This post has been edited 2 times. Last edited by popcorn1, Jul 20, 2021, 8:40 PM
,
.
move, color walnut 
brown walnuts between two green walnuts on either side. It's impossible to color all these walnuts green. To see this, note that if 
,
brown walnuts in total. Because this is odd, the groups are of different parities, so one group must have an even number of walnuts. Then keep splitting this group until you get a gap of two brown walnuts, at which point you can't color it.
brown walnuts into two groups. As 
walnuts 
for some 
.
and 
are the neighbours of 
,
.
,
moves, the number of obedient walnuts with labels greater than 
,
be the mentioned walnuts, in that order. If 
are less than 
either doesn't change or doesn't matter to us.
consecutive blue points, contradiction.
)
conservative if on the 
and 
with 
,
.
,
is not conservative, since it is swapped after the 
with 
that Bushy cannot take away any either walnut, a contradiction.
and 
.
.
.
)
remains constant. This can be asserted in a straightforward manner, which I will choose to omit here. This yields that in the end 
.
at the end. This is the contradiction which solves the problem.
t
at the beginning. The ![\[\sum_i |t_i - t_{i+1}|\]](http://latex.artofproblemsolving.com/5/4/b/54b486eb840758949053e21493abe9c04dfdf889.png)
changes by exactly 
-- after an odd number of moves, the sum should be 
!
.
has to be changed. When we change hole 
must be changed. This pattern continues, until we have that hole 
must be changed after hole 
.
.
,
is clear. Now assume that for an even 
'
.
,
is odd, implying that one of 
-
,
,
,
swapped during the process(call this a "forbidden move").
is trivial, so assume that 
work. Then on the first move of that chain, the chain will be split into two parts of length 
,
We want to then show that there is some swap that swaps a number 
and a number 
This is doable iff at every step, between every two consecutive 
walnuts there is an odd number of 
as we can only add a black walnut in between two black walnuts or in between two white walnuts. This is a contradiction as our final state with 
is odd, and after all moves are complete, all acorns are large, so 
.
be 
if both are smaller, and 
.
move. If 
,
,
.
,
.
