[/quote]
,
.
Let 
.
be positive real numbers such that 
.
.
is strictly increasing : 
satisfying the following conditions:
divides 
,
divides 
,
.
and 
. Prove that
be positive reals such that 
.![\[\frac{1}{a^5(b+2c)^2} + \frac{1}{b^5(c+2a)^2} + \frac{1}{c^5(a+2b)^2} \ge \frac{1}{3}.\]](http://latex.artofproblemsolving.com/3/b/d/3bd44cede90261ab949beabd32f8c08411ddc051.png)
such that for all 
,![\[
f(x + f(2y)) + f(x^2 - y) = f(f(x)) f(x + 1) + 2y - f(y).
\]](http://latex.artofproblemsolving.com/d/1/0/d10c01d7d416243614b24f04218d157e8b025ff2.png)
such that for all 
.
such that for all 
such that 
there exist 
such that 
and 
. Note : 
and 
for all 
. 
.
determine the minimum and maximum value of![\[\frac{\sum_{i,j=1}^n|a_i+a_j|}{\sum_{i=1}^n|a_i|}.\]](http://latex.artofproblemsolving.com/6/f/2/6f2427edf9a5e8e8f2e12276d0a841765e8322f1.png)
such that 
.
. 
,
(where 
)
for which ![\[s_ia_i+s_{i+1}a_{i+1}+\dots+s_ja_j=0.\]](http://latex.artofproblemsolving.com/4/b/6/4b66fe235e9a56cc646dbd38aae72f8ddfeb77de.png)
[/list]
Y by KevinYang2.71, peace09, nchuong1312, aidan0626
Determine the maximal length 
of a sequence 
of positive integers satisfying both the following properties:
of a sequence 
of positive integers satisfying both the following properties:
- every term in the sequence is less than or equal to
, and - there does not exist a consecutive subsequence
(where 
) with a choice of signs 
for which ![\[s_ia_i+s_{i+1}a_{i+1}+\dots+s_ja_j=0.\]](//latex.artofproblemsolving.com/4/b/6/4b66fe235e9a56cc646dbd38aae72f8ddfeb77de.png)
This post has been edited 1 time. Last edited by OronSH, Jul 17, 2024, 12:28 PM
.
.
that satisfies the conditions.
,
,
that satisfies this condition.
,
is possible. Clearly, the sequence 
satisfies this.
be a sequence that satisfies 
.
denoted as 
.
after 
.
is less than or equal to 
.
in 
,![\[
s_i \left( \frac{k_i}{2} \right) + s_{i+1} \left( \frac{k_{i+1}}{2} \right) + \dots + s_j \left( \frac{k_j}{2} \right) = 0
\]](http://latex.artofproblemsolving.com/5/d/3/5d3068b2970d37ea8c95b4625d7cd0e4bfe53f5a.png)
which contradicts the assumption from the induction hypothesis of case 
.
.
,
that satisfies the above conditions.
.
,
.
,
and for any integer 
where 
,![\[ b_i = s_1 a_1 + s_2 a_2 + s_3 a_3 + \dots + s_i a_i \]](http://latex.artofproblemsolving.com/b/6/5/b65bbccabf0290e9252431e2733a39e14656b428.png)
where 
,
for all integers 
where 
,
.
.![\[ -(2^{2023} - 1) \leq b_i \leq 2^{2023} \]](http://latex.artofproblemsolving.com/8/3/d/83d1b2d65b21e0ee14ec97f105c296e4eff74a00.png)
![\[ -(2^{2023} - 1), -(2^{2023} - 2), \dots, -1, 1, 2, \dots, 2^{2023} \]](http://latex.artofproblemsolving.com/6/6/d/66d3fab22c3b77d7ecd71cca42b875684f984ad5.png)
and 
are equal, where 
.![\[ b_t - b_s = 0 \Rightarrow s_{s+1} a_{s+1} + s_{s+2} a_{s+2} + \dots + s_t a_t = 0 \]](http://latex.artofproblemsolving.com/a/1/0/a10ac613679d43bfef7b3ab44773ef7b719f454a.png)
where 
is replaced by 
.
bad if there exist 
such that![\[a_is_i+a_{i+1}s_{i+1}+\ldots+a_js_j=0.\]](http://latex.artofproblemsolving.com/0/9/5/0959f5d1cc7febe019e889503712dc9995cac1f6.png)
The construction for 
is by induction: start with the sequence 
for 
and construct the sequence 
as 
(that is, 
,
,
)
doesn't work, we fix signs for each 
by focusing on the prefixes. That is, we start by picking 
depending on the previously picked 
,
,
.
.
is non-negative and strictly less than 
as 
.
with 
,
,
.
,
where we replace 
be a subsequence. Then there exists a suitable choice of signs so that 
is between 
and 
.
inductively. Let 
,
.
.
.
There are 2 cases.
.
.
.
is 
which obviously satisfies the condition and we're done.
.
does not satisfy the conditions.
.![$$[b_1,b_2],[b_3,b_4],...,[b_{2(\lfloor k/2 \rfloor)-1},b_{2(\lfloor k/2 \rfloor)}]$$](http://latex.artofproblemsolving.com/9/4/7/9479e12fe4e79b73ef67f70228d1ee992101a1c8.png)
If for any of the ranges the sum we get is 
we get by replacing each of the ranges with the value we get (and possibly removing the sequence that occurs after 
if ![$[b_2,b_3],[b_4,b_5],...$](http://latex.artofproblemsolving.com/9/d/3/9d3d0195b16f36e6e23765e12413fc3fdf450e8c.png)
(
and we are done by induction.
where 
.
and 
.
,
To see why this works, note that the subsequence cannot go through 
multiplied by 
and 
is clear. For the inductive step, the idea is to consider the partial sums 
(and 
)
of these, so by Pigeonhole, at least 
of these have the same parity. This means that we can split group the numbers into 
blocks so that each block has even sum. For example, if if 
and 
,
,
,
,
have the same parity, then the blocks are 
.
signs so that the results is in ![$[0, 2^{k+1}-2]$](http://latex.artofproblemsolving.com/a/6/7/a6760b4914dee1eecd9d967d87ce6aecee675ccd.png)
(by choosing one by one and maintain that interval). If any of these were ![$[1, 2^k-1]$](http://latex.artofproblemsolving.com/e/6/0/e600df0ad9b8da4fb6317ef129baf9c92ef237fd.png)
,
.
![\[ A_0=1 \]](http://latex.artofproblemsolving.com/1/9/6/1960cf1860a5eea7946850ba4d601907b08fc36c.png)
![\[ A_1=2, 1, 2 \]](http://latex.artofproblemsolving.com/7/f/e/7feb81b3a7d1380e45b5f36836254bf3df9235a3.png)
![\[ A_2=4, 2, 4, 1, 4, 2, 4\]](http://latex.artofproblemsolving.com/a/5/6/a56f3f22517de3a620d06b8633607385567200b3.png)
![\[ A_3=8, 4, 8, 2, 8, 4, 8, 1, 8, 4, 8, 2, 8, 4, 8 \]](http://latex.artofproblemsolving.com/d/6/a/d6aeac727c1772c388928845e557d4fcd4feb5e5.png)
![\[ \cdots \]](http://latex.artofproblemsolving.com/4/3/2/43216533b76d551f841a2b705f2986992ea1232d.png)
To get from 
to 
,
is our desired equality case for 
,
in 
.
has such a contiguous block. AFTOC that no such block exists. Consider a sequence 
with the following inductive definition![\[ e_k =
\begin{cases}
1 & \text{if } e_1a_1+e_2a_2+ \cdots e_ka_k \leq 0\\
-1 & \text{otherwise}
\end{cases} \]](http://latex.artofproblemsolving.com/f/2/7/f2729125430ac1289bfda800951ee49794b0b95e.png)
for all 
.
for any 
and 
then 
which contradicts our assumption. Thus, if we consider the set 
denoting all possible values of 
we see that 
and at least one of 
are not in 
implying by PHP that some 
appears at least ![\[ \left \lceil \frac{2^{n+1}}{|S|} \right \rceil \geq 2 \]](http://latex.artofproblemsolving.com/a/5/6/a56aaec37e6f2316b03bf691baf4b931f3018e63.png)
times in ![\[ e_1a_1+\cdots +e_ia_i = e_1a_1+\cdots +e_ja_j = d \]](http://latex.artofproblemsolving.com/2/d/c/2dc339ffd13455bfe048b93929abd5f075d1bdf1.png)
but then, ![\[ e_{i+1}a_{i+1}+ \cdots +e_ja_j =0 \]](http://latex.artofproblemsolving.com/f/f/b/ffb80c71cfc57b590d0310b38fe0f16493ee47a7.png)
Contradicting our assumption. 
for general 
.
,
.
or 
,
not being hard to see as the maximal sequence is 
.
positive integers less than 
,
has a subsequence that can be made into 
where 
is an increasing sequence. We can combine terms 
,
.
instead and now this is guaranteed to have at least 
and consider the subsets 
for all 
recursively. For 
just pick 
.
for 
we will copy the first 
if 
and 
if 
(notice that if 
we are done) so the sum for 
will be 
.
for all 
)
with 
.
are equal to those of 
so in fact we have:
just take 
.
which satisfies the problem for 
and there doesn't exist a choice of signs for it by the inductive hypothesis.
,
.
is impossible, consider starting from ![$[-(2^{2023}-1),2^{2023}]$](http://latex.artofproblemsolving.com/1/3/a/13ae546d15fcae88b3b52ce8ff2ac2d4cb8d7c97.png)
.
nonzero numbers in this interval, contradiction. 
,
for 
and 
for 
as follows:
for 
for some 
.
for all 
.
,
,
.
,
,
,
.
.
for 
.
,
such that the sum 
.
,
,
,
.
,
,
,
.
,
,
as just number 
as sequence 
and 
.
the sum can't be zero. On the other hand, the parts on the 
and 
of 
.
terms, say 
.
.
.
has a sequence of '+' and '-' that we searched for. Then the sequence 
and 
these signs will be different (that form 
