chứng minh

G

goku123123

đặt a-b=x;b-c=y;c-a=z
Ta có
$\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}$
=$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$
=$(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2-2(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz})$
=$(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2-\frac{2(x+y+x)}{xyz}$
=$(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2-\frac{2(a-b+b-c+c-a)}{xyz}$
=$(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2$
=$(\frac{1}{a-b}+\frac{1}{b-y}+\frac{1}{c-a})^2$
\Rightarrow thuộc Q

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