This question already has an answer here:
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 $2S=2(xy+zx+zy)=(x+y+z)^2 -x^2-y^2-z^2=(4-y)^2-x^2-y^2-z^2$
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 ??