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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 ??


marked as duplicate by Martin R, Community Apr 18 at 7:19

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  • $\begingroup$ Are $$x,y,z$$ assumed to be positive? $\endgroup$ – Dr. Sonnhard Graubner Apr 18 at 5:32
  • $\begingroup$ $4$ is the right result. $\endgroup$ – Dr. Sonnhard Graubner Apr 18 at 5:35
  • $\begingroup$ @MartinR The OP has a solution and asks for a confirmation. Not a dupe. $\endgroup$ – Jean-Claude Arbaut Apr 18 at 7:10
  • $\begingroup$ @Jean-ClaudeArbaut: You have a point. But the “... is there a better way” part has been answered before. Btw. none of the present answers addresses the “is this right” part of the question. $\endgroup$ – Martin R Apr 18 at 7:17

Using $$\color{red}{ab\leq \frac{1}{4}(a+b)^2}\;\forall\; a,b \in \mathbb{R}$$

Equality hold when $\color{red}{a=b}$

So $$(x+y)(y+z)\leq \frac{1}{4}\bigg[(x+y)+(y+z)\bigg]^2$$

$$xy+yz+zx+y^2\leq 4\Rightarrow \color{Red}{xy+yz+zx\leq 4-y^2\leq 4}$$

equality hold when $y=0$ and $x=z=2$


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 $$0\le (x-z)^2+4y^2$$


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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