cho a, b, c đôi một khác nhau. chứng minh: `a^2/(b-c)^2+b^2/(c-a)^2+c^2/(a-b)^2 ≥ 2`

Đánh giá :

$(\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b})^2\ge 0\\\to \dfrac{a^2}{(b-c)^2}+\dfrac{b^2}{(c-a)^2}+\dfrac{c^2}{(a-b)^2}+2 (\dfrac{ab}{(b-c)(c-a)}+\dfrac{bc}{(c-a)(a-b)}+\dfrac{ac}{(b-c)(a-b)})\ge 0(1)$

Xét :

$\dfrac{ab}{(b-c)(c-a)}+\dfrac{bc}{(c-a)(a-b)}+\dfrac{ac}{(b-c)(a-b)}$

$=\dfrac{ab (a-b)+bc(b-c)+ac(c-a)}{(a-b)(b-c)(c-a)}\\=\dfrac{ab(a-b) + b^2c - bc^2 + ac^2-a^2c}{(a-b)(b-c)(c-a)}\\=\dfrac{ab (a-b) - c (a-b)(a+b) +c^2 (a-b)}{(a-b)(b-c)(c-a)}\\=\dfrac{(a-b)(ab - ac-bc +c^2)}{(a-b)(b-c)(c-a)}\\=\dfrac{(a-b)[a (b-c) -c (b-c)]}{(a-b)(b-c)(c-a)}\\=\dfrac{-(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}\\=-1\\(1)\to \dfrac{a^2}{(b-c)^2}+\dfrac{b^2}{(c-a)^2}+\dfrac{c^2}{(a-b)^2} -2\ge 0\\\to \dfrac{a^2}{(b-c)^2}+\dfrac{b^2}{(c-a)^2}+\dfrac{c^2}{(a-b)^2} \ge 2$