find the maximum value of $xy + yz +zx$ find the maximum value of $$xy + yz +zx$$given that $x+2y+z=4$
my attempt :
$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx) $ 
or $2S=2(xy+zx+zy)=(x+y+z)^2 -x^2-y^2-z^2=(4-y)^2-x^2-y^2-z^2$
$2S=-x^2-z^2-8y+16=-x^2-z^2+4x+4z$
from the the above we can say due to symmetry maximum value occurs at $x=z$
hence $S=-x^2+4x$ whose maximum is 4
is this right or/and is there a better way ??

We have $$\frac{(x+2y+z)^2}{16}=1$$ so we have to prove that
$$4xy+4yz+4zx\le x^2+4y^2+z^2+2xz+4yz+4xy$$ and this is $$0\le (x-z)^2+4y^2$$
