$xy+yz+zx=xyz$ => $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$
Đặt $\frac{1}{x}=a, \frac{1}{y}=b, \frac{1}{z}=c$   $(a,b,c>0)$
 $A=b^2(\frac{1}{a}-1)+c^2(\frac{1}{b}-1)+a^2(\frac{1}{c}-1)$
$=(\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b})-(a^2+b^2+c^2)$
$=(a+b+c)(\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b})-(a^2+b^2+c^2)$  (do a+b+c=1)
$=(\frac{a^3}{c}+\frac{b^3}{a}+\frac{c^3}{b})+(\frac{a^2b}{c}+\frac{b^2c}{a}+\frac{c^2a}{b})$
Theo BĐT Cauchy: 

$\frac{a^3}{c}+ca \geq 2a^2$

$\frac{b^3}{a}+ab \geq 2b^2$

$\frac{c^3}{b}+bc \geq 2c^2$

=> $\frac{a^3}{c}+\frac{b^3}{a}+\frac{c^3}{b}+ab+bc+ca \geq 2(a^2+b^2+c^2)$

=> $\frac{a^3}{c}+\frac{b^3}{a}+\frac{c^3}{b} \geq 2(a^2+b^2+c^2)-(ab+bc+ca) $ (1)

$\frac{a^2b}{c}+bc \geq 2ab$

$\frac{b^2c}{a}+ca \geq 2bc$

$\frac{c^2a}{b}+ab \geq 2ca$

=> $\frac{a^2b}{c}+\frac{b^2c}{a}+\frac{c^2a}{b} \geq 2(ab+bc+ca)-(ab+bc+ca)=ab+bc+ca$ (2)

từ (1) và (2) => $A \geq 2(a^2+b^2+c^2)-(ab+bc+ca)+ab+bc+ca=2(a^2+b^2+c^2)$

theo BĐT Bunhia có: $(a+b+c)^2 \leq 3(a^2+b^2+c^2)$

=> $2(a^2+b^2+c^2) \geq \frac{2}{3}$

$A_{min}=\frac{2}{3}$ tại $a=b=c=\frac{1}{3}$ hay $x=y=z=3$