12. $$a^2+b^2+c^2=3abc \rightarrow \frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}=3$$
โดยอสมการ Cauchy-Schwarz ; $$(a+b+c)(\frac{a}{b^2c^2}+\frac{b}{c^2a^2}+\frac{c}{a^2b^2}) \geq (\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab})^2 =9$$
$$\therefore \frac{a}{b^2c^2}+\frac{b}{c^2a^2}+\frac{c}{a^2b^2} \geq \frac{9}{a+b+c}$$
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