# find the maximum value of $xy + yz +zx$ [duplicate]

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 $$2S=2(xy+zx+zy)=(x+y+z)^2 -x^2-y^2-z^2=(4-y)^2-x^2-y^2-z^2$$

$$2S=-x^2-z^2-8y+16=-x^2-z^2+4x+4z$$

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 ??

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• Are $$x,y,z$$ assumed to be positive? – Dr. Sonnhard Graubner Apr 18 at 5:32
• $4$ is the right result. – Dr. Sonnhard Graubner Apr 18 at 5:35
• @MartinR The OP has a solution and asks for a confirmation. Not a dupe. – Jean-Claude Arbaut Apr 18 at 7:10
• @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. – Martin R Apr 18 at 7:17

## 3 Answers

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 $$z=2,\,xy+yz+zx=4$$.