Prove the inequality.... 
Let $x,y$ be positive reals such that $x+y=2$. Prove that :
$x^3y^3(x^3+y^3) \leq 2$
Source : INMO 2002

My attempt :
I started with the left side of the inequality to be proved.
$x^3y^3(x^3+y^3) = x^3y^3(x+y)(x^2+y^2-xy) = 2 x^3y^3(x^2+y^2+2xy-3xy)$
$=2x^3y^3(4-3xy)$
How to proceed ?
Do I have to some AM-GM or Cauchy-Schwarz on particular set of values ?

You are nearly done.
Applying AM-GM gives us that $$  xy \times xy  \times xy \times (4-3xy) \le  \frac {(4-3xy+xy+xy+xy)^4}{4^4}=1$$
