AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be positive real numbers.![$$\sqrt[3]{\left(\frac{a^{2022}-a}{b}+\frac{2021}{a^{\frac{a}{b}}}+1\right)\left(\frac{b^{2022}-b}{c}+\frac{2021}{b^{\frac{b}{c}}}+1\right)\left(\frac{c^{2022}-c}{a}+\frac{2021}{c^{\frac{c}{a}}}+1\right)}$$](http://latex.artofproblemsolving.com/c/4/9/c4923b16be8a8faf86422477a6ed6b4b60eb2552.png)
satisfying ![$$(ab+bc+ca)\left[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\right]\ge \frac{9}{4}+k\cdot\frac{a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)}{(a+b+c)^{3}}$$](http://latex.artofproblemsolving.com/f/6/e/f6ed10f7fff1cc94edd8f451e75718a0916a8bfa.png)
holds for all 
+1 w
black and 
red cards. You pull out 
cards, one after another, and check their colour. If both cards are the same colour, then a black card is added to the deck. However, if the cards are of different colours, then a red card is used to replace them. Once the cards are taken out of the deck, they are not returned to the deck, and thus the number of cards keeps reducing. What is the likelihood the last card left in the deck is black?
be positive real numbers such that 
Prove that we have![\[(a+b)(b+c)(c+a)\geq 8.\]](http://latex.artofproblemsolving.com/2/a/6/2a63f19de634fce70ff4ebc02ad13d442a20c378.png)
such that the equality 
holds for all pairs of real numbers 
.
such that 
and 
.
be a given positive real and 
be the set of all positive reals. Find all functions 
such that ![\[f((c+1)x+f(y))=f(x+2y)+2cx \quad \textrm{for all } x,y \in \mathbb{R}_{>0}.\]](http://latex.artofproblemsolving.com/b/0/6/b069e7b7ec1e277f4a4ce85a99434fdb54eb02f3.png)
of the vertices of the regular polygon 
we write the number 
and in the remaining ones we write the number 
Let 
be the number written on the vertex 
A vertex is good if ![\[x_i+x_{i+1}+\cdots+x_j>0\quad\text{and}\quad x_i+x_{i-1}+\cdots+x_k>0,\]](http://latex.artofproblemsolving.com/b/f/4/bf4cf984af7f310c1eb90b6b37126bfca9181305.png)
for any integers 
and 
such that 
Note that the indices are taken modulo 
Determine the greatest possible value of such that, regardless of numbering, there always exists a good vertex.
be a scalene triangle, 
be the median through 
, and 
be the incircle. Let touch 
at point 
and segment 
meet for the second time at point 
. Let 
be the triangle formed by lines and and the tangent to at . Prove that the incircle of triangle is tangent to the circumcircle of triangle .
of positive integers satisfying the following two conditions:
, if 
is in , then so are 
and 
; and contains an integer 
such that 
is not divisible by 
.
of positive integers satisfying the following two conditions:
, if is in , then so are and ; and contains an integer such that is not divisible by .
contains least one element that is divisible by and at least 
.
, then the statement is trivial.
with 
. Since 
, we must have 
and 
both in . If is even, then either or are divisible by and at least 
, satisfying the condition. Otherwise, both of them are odd, and one of them must be congruent to 
modulo . In that case, we again find an element in that is divisible by and at least .
and 
, then 
.
. Then 
, which implies .
.
. Now, suppose 
for some 
. Then 
. Since 
, we can repeat this process indefinitely to generate larger and larger multiples of 8 in . This implies that contains arbitrarily large multiples of . Hence, by Claim 2, the desired result follows.
.
for any 
. Hence, 
.. (qed)
then 
where 
is "generally smaller" than 
and 
is "generally larger" than , so that you can go smaller and larger and possibly cover all of the domain.
, for some . More on this below.
impliying 
instead of 
and . The same steps apply in that case if 
is divisible by , however I haven't managed to prove it in general for all odd . If anyone has a solution for this modified version, I'd be interested to see it!
except for 
and 
(as verified by a computer search) but I don't think a clean general argument can be made.
is crucial here and here's why a general argument for that case couldn't exist. Going back to the second point above, it can be proven that any number 
can be written as a sum 
except for at most 
numbers. These are: 
. Note that all of them (except ) are mod . Now here's the most interesting part: there are 
numbers on that list and it's not known if the 19th exists. If the Generalized Riemann Hypothesis is true, then there is no 19th number. See here for more details.
and 
Prove that
is an obvious solution. Now, let 
.
or 
.
.
