[/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 
.
be the points on 
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
Let 
be an integer. Determine the smallest positive integer 
such that some cells on an 
board can be painted black so that in each row and in each column there are exactly 
black cells, and furthermore, the black cells do not share a side or a vertex with another black square.
Clarification: You have to answer n based on.
be an integer. Determine the smallest positive integer 
such that some cells on an 
board can be painted black so that in each row and in each column there are exactly 
black cells, and furthermore, the black cells do not share a side or a vertex with another black square.Clarification: You have to answer n based on
.
