WLOG, giả sử $ x \ge y \ge z$
$\dfrac{1}{(x-y)^2} + \dfrac{1}{(y-z)^2} + \dfrac{1}{(z-x)^2} = \dfrac{1}{(x-y)^2} + \dfrac{(y-z)^2 + (z-x)^2}{(y-z)^2(x-z)^2}$
$ = \dfrac{1}{(x-y)^2} + \dfrac{y^2 -2yz + 2z^2 + x^2 -2xz}{(y-z)^2(x-z)^2}$
$ = \dfrac{1}{(x-y)^2}+ \dfrac{(x^2 -2xy +y^2) + 2(z^2 -xz - zy +xy)}{(y-z)^2(x-z)^2} $
$ = \dfrac{1}{(x-y)^2}+ \dfrac{ (x-y)^2 + 2(z-x)(z-y)}{(y-z)^2(x-z)^2}$
$ = \dfrac{1}{(x-y)^2}+ \dfrac{(x-y)^2}{(y-z)^2(x-z)^2} + 2 *\dfrac{1}{(z-y)(z-x)}$
Ta có $a^2 +b^2 \ge 2ab\ ∀\ a;b $
$\to \dfrac{1}{(x-y)^2}+ \dfrac{(x-y)^2}{(y-z)^2(x-z)^2} \ge 2* \dfrac{1}{(z-y)(x-z)}$
$\to \dfrac{1}{(x-y)^2}+ \dfrac{(x-y)^2}{(y-z)^2(x-z)^2} + 2 *\dfrac{1}{(z-y)(z-x)} \ge \dfrac{4}{(z-y)(z-x)}$
Ta sẽ chứng minh
$ (z-y)(z-x) \le xy +yz + xz \to z^2 - zx -yz +xy \le xy + yz +xz$
$\to z^2 - zx - yz - xz \le 0 \to z(z-2x-y) \le 0$ (đúng vì $z \le y \le x$ )
BĐT đề bài được chứng minh
