prove that in every circle arrangment of {1,2,...n} there exists three consecutive elements (three neibours in the cycle arrangement) whose sum is at least [3(n+1)/2], (the upper integer value)
Thank's in advance
Forgive me if I have misunderstood something but I think the "trick" is to realize that the highest sum of three consecutive elements is n(n-1)(n-2)/3. And then you just need to prove that this sum is at least as big as 3(n+1)/2.
Suppose there is a circle arrangement
$\displaystyle a_1,a_2,a_3,\text{...},a_n,a_1,\text{...}$
for which this is not true. This means
$\displaystyle a_1+a_2+a_3\leq \left\lceil \frac{3(n+1)}{2}\right\rceil -1$
$\displaystyle a_2+a_3+a_4\leq \left\lceil \frac{3(n+1)}{2}\right\rceil -1$
$\displaystyle .$
$\displaystyle .$
$\displaystyle .$
$\displaystyle a_n+a_1+a_2\leq \left\lceil \frac{3(n+1)}{2}\right\rceil -1$
add them up
$\displaystyle 3\sum _{i=1}^n a_i\leq n \left\lceil \frac{3(n+1)}{2}\right\rceil -n$
$\displaystyle \frac{3n(n+1)}{2}\leq n \left\lceil \frac{3(n+1)}{2}\right\rceil -n$
$\displaystyle \frac{3(n+1)}{2}+1\leq \text{ }\left\lceil \frac{3(n+1)}{2}\right\rceil $
contradiction
I think this has been a greatly misunderstood problem.
Suppose that $N=12$ The question is about circular arrangements of which there are $11!$ possible.
Further of all those possible arrangements show that every one of them contains three neighbors $\displaystyle {a_k} + {a_{k + 1}} + {a_{k + 2}} \ge \left\lceil {\frac{{3(N + 1)}}{2}} \right\rceil $
In the case $N=12$ that number is $20$, so if $\displaystyle {a_k}=10, ~ {a_{k + 1}}=2 ,~\&~ {a_{k + 2}}=5 $ those neighbors add to $17$ so they do not work.
Look at Idea's solution to see a really innovative proof.
I point out to you that this is an English Language website. You are welcome here as long as you know that.
I fear that I am not sure that I understand what you wrote. That said take my example in which $N=12$ that is arranging the numbers $\{1,2,3,\cdots,11,12\}$ in a circular pattern: a random analog clock face. That are $11!=39,916,800$ ways to do that. This question asks you to prove the in each of those almost forty million arrangements there are three neighbors whose sum is at least $\left\lceil {\frac{{3(11 + 1)}}{2}} \right\rceil $. Actually you are asked to prove this in general not just for one value of $N$. Now maybe you miss-read the question written in English?
It can be successfully solved with the 'Principle Of Mathematical Induction'.
As we know the smallest circle can be formed with 3 positions(numbers in this case.)
Step 1:Prove for the smallest circle that it is true-
Sum of elements=3+2+1=6
which is =3(n+1)/2=6
So it is true for n=3;smallest circle
Step 2:Let it be true for n=m;
,i.e., m+(m-1)+(m-2)>=3(m+1)/2
=> 3m-3>=3(m+1)/2=(3m+3)/2
Step 3:Now prove that it will be true for n=m+1;
,i.e., (m+1)+m+(m-1)=3m
we need to prove that it greater 3(m+2)/2=(3m+6)/2=(3m+3)/2+(3/2)
In the LHS the sum increased by 3 whereas in the RHS only by 3/2. (as compared to n=m);
so,definitely this property is true for n=m+1
This way we can prove that for any two consecutive numbers it is true;i.e.,true for entire +ve integer system>3