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be a sequence of positive real numbers such that for every 
, we have:![\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]](http://latex.artofproblemsolving.com/e/3/0/e30e9e533eb9a41cf847484e676ac4522edda665.png)
Prove that there exists a natural number 
such that for all 
, the following holds:![\[
a_n < \frac{1}{1404}
\]](http://latex.artofproblemsolving.com/0/1/f/01f1a81eed2230332d51a15501ef2a6ad3a7ec82.png)
, points 
and 
lie on 
and 
respectively such that 
. Points 
and 
are the reflections of and with respect to the midpoints of sides and , respectively. Point 
lies on segment 
such that the circumcenter of triangle 
coincides with the circumcenter of triangle . passes through the midpoint of segment 
.
be the set of positive real numbers.
such that there exists a function 
with the property:![\[
f(y^2f(x) + y + c) = xf(x+y^2)
\]](http://latex.artofproblemsolving.com/e/3/0/e30ee6d9f69cc43422a5c2b9813192a632d359cb.png)
for all 
.
with no obtuse angles.![$$\frac1{\sqrt3}\cdot\left(\cot A+\cot B+\cot C\right)+\left(2-\sqrt 3\right)\sqrt[3]{\cot A\cot B\cot C}\geq\frac2{\sqrt3}.$$](http://latex.artofproblemsolving.com/d/1/4/d14154281d7ccf60a2dcfe1d937b974ae1d74f97.png)
such that : 
is non-increasing.
.
and 
gives![\[f((y-a)f(a)+\sqrt{y-a}+c)=af(y)<bf(y)=f((y-b)f(b)+\sqrt{y-b}+c)\]](http://latex.artofproblemsolving.com/f/e/6/fe6b840479f1eb218e9418aa9e165ba8aa137f3d.png)
for all 
.
.
,
gives LHS<RHS. Hence by IVT, such 
exists, contradiction. 
,
,
and 
,
.
,
.
and 
gives![\[af(a+y)=bf(b+y)\]](http://latex.artofproblemsolving.com/f/e/e/fee275f8872b721d726b266b56ba1b170c6ae7ce.png)
So for sufficiently large 
,
,![\[f(k(a-b)+r)=\left(\frac{b}{a}\right)^kf(r)\]](http://latex.artofproblemsolving.com/f/0/f/f0f76f5796fa93149b629665f52b79087296d182.png)
![\[\log f(k(a-b)+r)=\log f(r) - k(\log a - \log b)\]](http://latex.artofproblemsolving.com/8/2/0/8204717fddbbdaa57cb1d3bd442ec4be837b1cb5.png)
And similarly, we have![\[\log f(k(c-d)+r)=\log f(r) - k(\log c - \log d)\]](http://latex.artofproblemsolving.com/f/0/a/f0a66b4e6fc459bb3de67c09924a2c7e79f9977e.png)
But note that for all reals 
,
such that 
,![\[\log f(r) - p(\log a - \log b)=\log f(p(a-b)+r)\le \log f(q(c-d)+r)= \log f(r) - q(\log c - \log d)\]](http://latex.artofproblemsolving.com/3/b/d/3bda4f73665cbb920b8ca552b524f9dabcee8a74.png)
![\[ \frac{p}{q} \ge\frac{\log c - \log d}{\log a - \log b}\]](http://latex.artofproblemsolving.com/b/6/7/b671d04a5b26903be42cc8b41e46bbc52abfdf69.png)
![\[\frac{c-d}{a-b}+\epsilon \ge\frac{\log c - \log d}{\log a - \log b}\]](http://latex.artofproblemsolving.com/2/f/b/2fbba7b83531358f94b729fe60d461cb8d902429.png)
and taking 
gives 
.
.
,
.![\[\frac{y^2f(1)+y+c-(1+y^2)}{\log\left(\frac{y^2f(1)+y+c}{1+y^2}\right)}=k\]](http://latex.artofproblemsolving.com/5/7/3/573bd5365a8c5a0605d8ac8a2828a9a3dc3ba9e1.png)
for all 
except for finitely many values, as 
only holds for finitely many values. Then,![\[\lim_{y\rightarrow\infty}\frac{y^2f(1)+y+c-(1+y^2)}{\log\left(\frac{y^2f(1)+y+c}{1+y^2}\right)}=k\]](http://latex.artofproblemsolving.com/a/a/b/aab3866fbbc6e4f3872a157ce8b392f3b77ecc91.png)
If 
,
whereas the numerator tends to 
,![\[\lim_{y\rightarrow\infty}\frac{y+c-1}{\log\left(1+\frac{y+c-1}{y^{2}+1}\right)}=k\]](http://latex.artofproblemsolving.com/c/f/8/cf8ce8970303560d514ecebf499c47f78b1d3362.png)
But by L'Hôpital's rule,![\[\lim_{y\rightarrow\infty}\frac{(y^2+1)(y^2+1+y+c-1)}{y^2+1-2y(y+c-1)}=k\]](http://latex.artofproblemsolving.com/e/7/a/e7a5a5ebfba833736b77a628e535d3c7cf517825.png)
and the numerator has degree 
,
,