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.
prove that![\[ \frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{d^2} + \frac{d^2}{a^2}\geq\frac{a}{d} + \frac{b}{a} + \frac{c}{b} + \frac{d}{c}\]](http://latex.artofproblemsolving.com/0/0/4/004e7f148bbb0c797688f5195aa4002dbc316470.png)
If you know same tasks, link them please
be the Fermat point of a 
. Prove that the Euler line of the triangles 
, 
, 
are concurrent and the point of concurrence is 
, the centroid of .
(i know 
well). i want to study so if you guys have some easy or normal problems please send me
?
chess board, except for the 
central square, there is a checker. In one move, a checker can jump over adjacent checker (diagonally), and this adjacent checker (that was jumped over) is removed. Is it possible to come up with a sequence of moves, so that only one checker is left on the board?
be a quadrilateral such that the excircles 
and 
of triangles 
and 
touching their sides 
and 
respectively touch the extension of 
at the same point . The segment 
meets at point 
, and the line meets at and 
. Prove that one of angles 
and 
is right
with 
and 
.
?
that it is possible to form a triangle from segments of length 
. Prove that it is possible to reduce one of the numbers by 
times and obtain the numbers 
so that segments with lengths can also be formed into a triangle.
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)
For a positive integer 
, let 
be the set of bijective functions from 
to itself. For a pair of positive integers 
such that 
, and for a permutation 
, we say the pair is expanding for 
if 
(a) Is it true that for all integers 
, there exists so that the number of pairs that are expanding for permutation is less than 
? (b) Does there exist a positive integer 
and a permutation so that the number of pairs that are expanding for the permutation is less than 
?
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 
, and 
for all integers 
. Prove that 
.
that can be subtracted from both the numerator and denominator of the fraction 
, to get, after the reduction, the fraction of form 
, where 
are single digit numbers.
and hence 
.
or equivalently 
.
and of these four only 
divides 
.
and here 
indeed works.