2015 IMO contest problem: If$x,\,y,\,z$ are real numbers such that $x+2y+3z=6$ and $x^2+4y^2+9z^2=12$, evaluate $xyz$.

If$x,\,y,\,z$ are real numbers such that $x+2y+3z=6$ and $x^2+4y^2+9z^2=12$, evaluate $xyz$.

There is something that this Olympiad problem might trick us because it's obvious that $x^2,\,4y^2,\,9z^2$ are squares of $x,\,2y,\,3z$ and hence, one has reason to believe that the proper first step in solving this problem is to square the first given equation:

$x+2y+3z=6$