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.
satisfying 
, find the biggest value the following equation could acquire:
+1 w![\[\frac{x^3+1}{(y-1)(z+1)}+\frac{y^3+1}{(z-1)(x+1)}+\frac{z^3+1}{(x-1)(y+1)}\]](http://latex.artofproblemsolving.com/5/c/6/5c6894b617b990ec69c62ed1cf7d062d4660af53.png)
where 
are reals.
so that for any 
, we have ![\[ f(xf(x^2)+yf(x))=xf(y+xf(x)). \]](http://latex.artofproblemsolving.com/1/d/4/1d4fbf3f80306aaae5cfc11cc0e4be7ea41491d7.png)
such that a positive integer power of 
equals to a positive integer power of 
.
positive real numbers satisfying 
,show that:![$\sqrt[3]{\frac{a^3+b^3+c^3}{3}}+\frac{a+b+c}{9} \geq \frac{4}{3}$](http://latex.artofproblemsolving.com/c/b/3/cb39e7f838c99ed8a209c6844f3d2810838d1dfd.png)
.
, where 
and 
. Prove that 
couldn't be expressed as product of two non-constant polynomials with integer coefficients.
be positive real numbers .Prove that
Prove that
Where 
Prove that
Let 
Prove that
Let 
Prove that
robots, each in a different cell. for which Alice can always prevent Bob from winning, no matter where he puts the robots.
and 
be prime numbers. Prove that if 
is a perfect square, then 
is also a perfect square.
in order to make it a perfect square?
be mid-points of 
respectively 
and since, 
and 
are isosceles 
is a kite,
and we also know, 
is isosceles
is altitude of 
,such, 
.........Let 
Therefore, 
, 
, 
, 
. Thus Midpoint of 
and midpoint of 
. Using distance formula we get 
.
such that 
are collinear. Let 
be the midpoint of 
, thus it is also the midpoint of 
. Now 
and 
. By pythagorean theorem, 
. And using MPT, the required answer is .
. Prove that
be an acute angled triangle and 
be the altitude through 
.
and 
.
be the midpoint of 
)
be the intersection of the line and 
.
and 
,
and 
.
,
.
using 
and 
.
,
and 
be the midpoints of 
on 
,
.
by the Midpoint Theorem.
,
.
is a right triangle.