# Find the maximum and minimum values of $A=\frac{y}{x}-\frac{x}{y}+\frac{z}{y}-\frac{y}{z}+\frac{x}{z}-\frac{z}{x}$

Find the maximum and minimum values of

$$A=\frac{y}{x}-\frac{x}{y}+\frac{z}{y}-\frac{y}{z}+\frac{x}{z}-\frac{z}{x}$$

$x,y,z$ are possitive real numbers satisfying $M\le4m$

where $M=max{(x,y,z)}$, $m=min{(x,y,z)}$

My try: Let $\frac{y}{x}=a,\frac{z}{y}=b,\frac{x}{z}=c$ so $abc=1$

$A=a+b+c-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}=a+b+c-ab-bc-ca=(a-1)(b-1)(c-1)$

$minA$ (may be) $=0$

And I don't know what to do more.

Thanks

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From $M \leq 4m$, you get $1/4 \leq a,b,c \leq 4$. And $abc=1$ implies atleast one of $a-1$, $b-1$, $c-1$ is non-positive and atleast one is non-negative. The third one can be either positive or negative (or zero).
For minimum value, without loss of generality, assume $a-1$ is negative and $b-1$, $c-1$ are positive. For any fixed $a<1$, $A$ is minimum when $(b-1)(c-1) = bc + 1 -(b+c)$ is maximum. As $bc$ is fixed, maximum is achieved when $b=c$. So for a fixed $a$, minimum value of $A$ is $(a-1)(\frac{1}{\sqrt{a}}-1)^2$. This can be seen to be strictly increasing for $a \in (0,1]$. So minimum value is at $a=1/4$.
Maximum value of $A$ can be proved similarly, or can be obtained directly by noticing that maximum will exactly be negative of minimum (just replace $a,b,c$ by $\frac{1}{a},\frac{1}{b},\frac{1}{c}$).
I think max and min values are $\pm \frac{3}{4}$. –  polkjh Jan 18 '13 at 13:26