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 
.
such that the equation 
has a rational root, if:
;
?
is a root for coprime 
;
.
are odd, both 
.
is odd, the sum on the LHS consists of odd number of odd terms, hence it's also odd.
for some prime 
works. Indeed, split the divisors of 
for 
and observe that 
has root 
.
is a root of the polynomial.