<s>[INST] 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 ??
 [/INST] With a Lagrangian multiplier $\lambda$ in a Lagrangian $L:=xy+yz+zx+\lambda (4-x-2y-z)$, $0=\partial_x L=y+z-\lambda$ etc. gives $\lambda=y+z=\frac{z+x}{2}=x+y$. Comparing the first and last of these expressions for $\lambda$ gives $x=z$, so $y+z=z$ and $y=0$. Then $x=\frac{4-2\times 0}{2}=2$, so $z=2,\,xy+yz+zx=4$.
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