Let $x,y,z$ be positive numbers such that $xyz=1$. Determine (with proof) extreme values of ${xy+yz+zx\over x+y+z}$

Question

Let $x,y,z$ be positive numbers such that $xyz=1$. Determine (with proof) extreme values of $${xy+yz+zx\over x+y+z}$$

Please, no calculus are allowed!

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Mark with $E$ extremum of that expression. So

$$E= {x^2y^2 +y+x\over (x+y)xy+1}$$

Then we have $$x^2y^2(1-Ey)+x(1-Ey^2)+y-E=0$$ which discriminat (on variable $x$) must be $0$, so

$$ (1-Ey^2)^2-4(y-E)(y^2-Ey^3)=0$$

so $$E^2(y^4-4y^3)+E(4y^4+2y^2)+1-4y^3=0$$

and here I'm stuck... Any help?

Answer 1

For $$ (x,y,z) = \left(n,n, \frac{1}{n^2} \right) $$ I got the requested ratio as $$ n \left(\frac{n^3 + 2}{2 n^3 + 1} \right) \approx \frac{n}{2} $$

For $$ (x,y,z) = \left(\frac{1}{n},\frac{1}{n}, n^2 \right) $$ I got the requested ratio as $$ \frac{1}{n} \left(\frac{2n^3 + 1}{ n^3 + 2} \right) \approx \frac{2}{n} $$

Answer 2

$$\frac{xy+xz+yz}{x+y+z}=\frac{xy+xz+yz}{(x+y+z)\sqrt[3]{xyz}},$$ which is homogeneous and we can ignore the condition already.

We see that $\frac{xy+xz+yz}{(x+y+z)\sqrt[3]{xyz}}\rightarrow+\infty$ for $z\rightarrow0^+$ and $x=y=1$ and $$\frac{xy+xz+yz}{(x+y+z)\sqrt[3]{xyz}}\rightarrow0$$ for $x=y=1$ and $z\rightarrow+\infty.$

Answer 3

Ahh, yes, (note that $x$ and $y$ run independently across positive real numbers) let $$f(y) = \lim_{x\to \infty} {x^2y^2 +y+x\over (x+y)xy+1} = \lim_{x\to \infty} {x^2(y^2 +{y\over x^2}+{1\over x})\over x^2(y+{y^2\over x}+{1\over x^2})} = y$$

Since $$\lim_{x\to \infty} f(y) = \infty$$ we have $E$ is unbounded. Similary we get

$$g(y) = \lim_{x\to 0} {x^2y^2 +y+x\over (x+y)xy+1} = y$$

so $\inf E =0$ and it is not minimum.