[/quote]
,
be the set of non-negative integers and 
be the set of positive real numbers. Let 
be a function such that 
and 
for all integers 
,
for all integers 
.
.
and 
Prove that
Let 
and 
Prove that
which satisfy 
![$ \frac{a+b+c}{3} \ge \sqrt[10]{\frac{a^3+b^3+c^3}{3}}$](http://latex.artofproblemsolving.com/c/d/a/cdae05ad34239366b0dd3921243522c711b6ae58.png)
method, I can assum 
![$ \frac{2x+\frac{1}{x^2}}{3} \ge \sqrt[10]{ \frac{2x^3+ \frac{1}{x^6}}{3}} $](http://latex.artofproblemsolving.com/c/d/4/cd481672bee1e87d8c97ac623436e0ec75cfa73a.png)
are given such that 
.
in natural numbers m and n.
,
denote the smallest possible value of 
where 
are sets such that 
and 
whenever 
.
and 
intersect at 
.
of the two circles touches 
at 
respectively. Let 
be the symmetric points of 
intersects circles 
respectively which are distinct from 
and 
are equal.
,
.
intersects 
at 
.
cuts segment 
at 
respectively. If 
and 
,
equals 
.
are points on a circle 
,
,
.
at 
.
is on 
such that 
.
intersects 
.
is the reflection of 
wrt 
.
lies on the same circle.
in its exterio
so that 
,
,
.
and
such that 
and 
.
and 
,
and 
.
and 
is on the circumference of triangle 
has a right angle at 
.
such that 
and 
lies between 
.
is chosen so that 
and 
.
is chosen so that 
and 
is the bisector of 
.
be the point such that 
is a parallelogram. Prove that 
and 
are concurrent.
such that the number of 
which satisfy the following condition is minimal:
has periodic 
and 
are tangent to the circumcircle of 
be the reflection of 
be the reflection of 
.
is the circumcenter of 
,
is perpendicular to 
Y by Adventure10, Mango247
Ana and Beto play against each other. Initially, Ana chooses a non-negative integer 
and announces it to Beto. Next Beto writes a succession of 
numbers, 
of them equal to 
and of them equal to 
. Once this is done, Ana must split the succession into several blocks of consecutive terms (each term belonging to exactly one block), and calculate the sum of the numbers of each block. Finally, add the squares of the calculated numbers. If this sum is equal to , Ana wins. If not, Beto wins. Determine all values of for which Ana can ensure victory, no matter how Beto plays.
and announces it to Beto. Next Beto writes a succession of 
numbers, 
of them equal to 
and of them equal to 
. Once this is done, Ana must split the succession into several blocks of consecutive terms (each term belonging to exactly one block), and calculate the sum of the numbers of each block. Finally, add the squares of the calculated numbers. If this sum is equal to , Ana wins. If not, Beto wins. Determine all values of for which Ana can ensure victory, no matter how Beto plays.
.
,
.
to 
,
.
or 
blocks, such that the sum in each block is either 
blocks all summing to either 
There exists 
consecutive blocks, 
and 
that sum to 
are of the same sign, which implies that
impossible as
If 
,
and combine them into one block. Otherwise, take 
and combine them into one block. We have found a new sequence of 
