[/quote]
,
such that for any positive integers 
we have 
.
,
is divisible by 
.
be a set of 2025 positive real numbers. For a subset 
,
as the median of 
when all elements of 
.![\[
P(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ odd}}} M_T, \quad Q(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ even}}} M_T.
\]](http://latex.artofproblemsolving.com/1/d/f/1df2ea8487289111939a53dcc49576fbe6fdef37.png)
Find the smallest real number 
such that for any set ![\[
P(A) - Q(A) \leq C \cdot \max(A),
\]](http://latex.artofproblemsolving.com/c/d/e/cdea343d11c722a423e857aeb01202863405773e.png)
where 
denotes the largest element in 
be an integer. Two players, Alice and Bob, play the following game on the complete graph 
:
,![\[
\left( \frac{1}{4} - \varepsilon \right) n^2 + n.
\]](http://latex.artofproblemsolving.com/a/9/c/a9cca00e87c643af32eb75a2501b59060f3504fb.png)
with 
.
takes place for all choiches of the 
signs. Show that there exists a line 
through 
such that all points 
are all on one side of 
of the form 
with 
,
Determine all ordered pairs of polynomials 
with integer coefficients such that 
.
be an integer. Show that, if 
is an integer, then it is a perfect square.[/quote]
.
is at least rational, so that 
must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs 
for which this happens are 
and 
,
,
The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?
such that![\[(a^2-b^2)(b^2-c^2) = abc.\]](http://latex.artofproblemsolving.com/7/1/8/7188261c40a45f5e15533100432d36abbf515efd.png)
.
that 
is a prime.
interesting if 
of 
and 
is also a divisor of 
angle 
.
is the midpoint of 
,
is the foot of altitude from point 
.
and 
are marked such that 
is equilateral, and 
and this points lie in the same half-plane with respect to 
and 
are tangent.
?
and 
are given in space, such that 
.
and 
be the midpoints of 
,
and 
.
is a right triangle;
and 
are perpendicular.
,
on 
,
be points on lines 
,
,
,
,
concur at a point 
be a point on line 
,
,
be a point on line 
be a point on line 
,
,
concur at a single point.
,
the sum of its digits. Find all integers 
such that there exist 
with 
for all 
.
Y by mathmaths, Snakes, caillouhascancer, jam10307, rightways, 62861, doxuanlong15052000, aopser123, Davi-8191, kk108, tenplusten, Davrbek, m2017m, centslordm, mathematicsy, donotoven, megarnie, Lamboreghini, vic_52math, sabkx, Qiqgloe, Adventure10, Mango247, NicoN9, cubres
Suppose that a sequence 
of positive real numbers satisfies ![\[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]](//latex.artofproblemsolving.com/0/3/9/03909a5f60d944063182ec97cf73f283178e8d4d.png)
for every positive integer 
. Prove that 
for every 
.
of positive real numbers satisfies ![\[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]](http://latex.artofproblemsolving.com/0/3/9/03909a5f60d944063182ec97cf73f283178e8d4d.png)
for every positive integer 
. Prove that 
for every 
.
,
.![\[ \frac{k}{a_{k+1}} \le a_k + \frac{k-1}{a_k}. \]](http://latex.artofproblemsolving.com/6/e/d/6edbf45db334736dd8d749161d86327ee5288dac.png)
Summing these all together, we now obtain a telescoping sum, and deduce the key fact ![\[ s_k \ge \frac{k}{a_{k+1}} \iff a_{k+1} \ge \frac{k}{s_k}. \]](http://latex.artofproblemsolving.com/6/b/0/6b0c96530983d239fc2c763f6da361bb38262419.png)
Now the result follows by induction: for any ![\[ s_{n+1} = s_{n} + a_{n+1} \ge s_n + \frac{n}{s_n} \ge n+1 \]](http://latex.artofproblemsolving.com/a/9/c/a9ca477be2e3227d2f407e4077441c2da202d09f.png)
where in the last step we use the fact that 
.
to 
,
.
following from 
.
since 
.
.
,
,
,
for all 
'
,
.
,
to 
gives![\[\sum_{1 \le i < j \le n}a_ia_j \ge \frac{n(n - 1)}{2}.\]](http://latex.artofproblemsolving.com/f/d/b/fdb401f17105d4f61a3725b08f4b093d6cb87aa6.png)
It is easy to see that 
from here.
for all 
.
where the first inequality is just Cauchy-Schwarz. 
.
.
![\[s_n + \frac{n}{s_n} \ge n+1 \]](http://latex.artofproblemsolving.com/e/1/3/e1332546219014129a8fbd4c87b502dfb217e9fe.png)
is strictly increasing in the interval 
,
,
then 
.
,
.
is positive for 
.
,
is equivalent to 
,
,
factors as 
it is clear that this holds for 
and show that 
since 
.
.
.
,
:
is trivial since 
.
.
:
which is true since 
.
where it is true by 
and in the condition if 
Suppose that it is true, then we prove that 
where 
for 
Then 
We know if this inequality 
then 
But it isn't true and this is contradiction.
) First, we take the base case 
.
Now we take the reciprocal of the inequality and multiply by 
Plugging in values of 
to 
Add all these inequalities to obtain 
where 
denotes 
.
.
We know 
.
![\[a_1+a_2 \geq a_1+\frac{1}{a_1} \geq 2\]](http://latex.artofproblemsolving.com/3/9/b/39befac47a5aabcc692059a24c59e47a5c234131.png)
Which follows by (AM-GM or rearrangement inequality).
.![\[\frac{1}{a_{k+1}} \le \frac{a_k^2+(k-1)}{ka_k} \implies \frac{k}{a_{k+1}} \le \frac{a_k^2+(k-1)}{a_k} = a_k + \frac{k-1}{a_k}.\]](http://latex.artofproblemsolving.com/5/f/0/5f055c648edde70b67f0782152640db1a77caf11.png)
So we have that![\[\frac{k}{a_{k+1}} - \frac{k-1}{a_k}. \leq a_{k+1}\]](http://latex.artofproblemsolving.com/9/3/3/93305184dfbde1ec1528c9ece8e5726a6a82db6a.png)
Which means that.![\[\sum_{k=1}^{n+1} \frac{k}{a_{k+1}} - \frac{k-1}{a_k} = \frac{n}{a_{n+1}} \]](http://latex.artofproblemsolving.com/6/3/e/63e6cacca6dcc0e898fd659d7949ca2f81cec560.png)
So we have that.![\[\frac{n}{a_{n+1}} \le a_1 + a_2 + \ldots + a_n \implies a_{n+1} \geq \frac{n}{\sum_{k=1}^{n+1}a_k}\]](http://latex.artofproblemsolving.com/8/a/3/8a38695cad82bf46edb29a7db4ffffcf1420fb12.png)
But this means that by our inductive hypothesis 
)![\[a_1+a_2+....a_{n+1} \geq a_1+a_2+\cdots+a_n+\frac{n}{a_1+a_2+\cdots+a_n} =
\frac{(a_1+a_2+\ldots +a_n)^2+n}{a_1+a_2+\ldots +a_n} \]](http://latex.artofproblemsolving.com/1/0/4/104a4845fbfaddc4001795327a9de98a4558b6a3.png)
![\[ \geq \frac{(n+1)(a_1+a_2+\ldots +a_n)}{a_1+a_2+\ldots + a_n}=n+1 \]](http://latex.artofproblemsolving.com/1/e/6/1e65d894f42a5870d6b59416176ce47561aed207.png)
In which the last inequality 
)
Rearranging the final expression also yields 
.![\[s_{k + 1} = s_k + a_{k + 1} \geq s_k + \frac{k}{s_k} \geq k + 1, \]](http://latex.artofproblemsolving.com/9/a/9/9a9ba6e551897347dc701ab95dd676c293649ad6.png)
as 
,
.
with induction on 
,
by AM-GM.
Adding this up from 
,
.
which holds since 
![\[\implies\frac{1}{a_{k+1}}\leq\frac{a_k^2+(k-1)}{ka_k}\implies\frac{k}{a_{k+1}}\leq a_k+\frac{k-1}{a_k}\]](http://latex.artofproblemsolving.com/a/7/e/a7e564f1957304080f97c1453c7b4cc1e42a3fd5.png)
![\[\implies a_k\geq \frac{k}{a_{k+1}}-\frac{k-1}{a_k}\]](http://latex.artofproblemsolving.com/5/5/6/55614adfc33d6b829f9806908ff223c7338f5507.png)
![\[\implies \sum_{i=1}^{n}a_i\geq\frac{n}{a_{n+1}}\implies a_{n+1}\geq\frac{n}{S(n)}\cdots(i)\]](http://latex.artofproblemsolving.com/6/1/6/616fe5aeb3d12c681ee6150085edee4079c131c4.png)
Let 
(by AM-GM).
,
Therefore, the claim is also true for 
and by the principle of mathematical induction, we're done!
.
and 
and 
and that 
in general, which is easy to prove by induction.
,
,
for some 
,
which finishes our induction, as desired.
by AM-GM. For the inductive step assume the claim is true for some 
immediately implies the desired conclusion so allow 
.![\[a_k \ge \dfrac{k}{a_{k + 1}} - \dfrac{k - 1}{a_k} \implies a_1 + a_2 + \dots + a_{n + 1} \ge a_{n + 1} + \dfrac{n}{a_{n + 1}} \]](http://latex.artofproblemsolving.com/4/1/7/41732e3e1a4943dbb1db6a3d5133e313cfbb4e99.png)
but![\[(a_{n + 1} - 1)(a_{n + 1} - (n + 1)) \ge 0 \implies a_{n + 1} + \dfrac{n}{a_{n + 1}} \ge n + 1\]](http://latex.artofproblemsolving.com/1/3/4/134ea81d3b478e89ccbfdf15dfa81d79d56842c9.png)
and we are done.
.
holds true, so let 
.
.
is telescoping and thus 
or 
.
,
,
.
for all 
,
,
for all 
,
and noting that the first derivative of 
is 
is positive for all 
for all ![\[(k-1)b_k\le\frac1{b_{k-1}}+(k-2)b_{k-1}\]](http://latex.artofproblemsolving.com/2/9/0/2900d4a81b5fd0c70ea3b84df9d8922af3f47f94.png)
for all 
.
for 
.
,
to 
and adding 
implies that 
since 
.
Induction Hypothesis: Suppose that 
for all 
,
,
.
Obs: Note that we could divide 
because if 
,
Let 
therefore adding everything for 
yields 
for all positive integers 
is trivial as 
by AM-GM.
observe that 
Now 
which is clearly true by the inductive hypothesis, so the proposition holds for 
Thus our induction is complete, and we are done. QED
we have 
.
.
and prove it for 
.
.
it suffices to show 
.
we have 
.
to both sides yields 
.
.
by AM-GM inequality.
for some 
.
