Ta có :$P=\frac{ab}{(b-c)(c-a)}+\frac{bc}{(c-a)(a-b)}+\frac{ca}{(a-b)(b-c)}=\frac{ab(a-b)+bc(b-c)+ca(c-a)}{(a-b)(b-c)(c-a)}=\frac{ab(a-b)-bc(a-b+c-a)+ca(c-a)}{(a-b)(b-c)(c-a)}=\frac{(a-b)(ab-bc)+(c-a)(ca-bc)}{(a-b)(b-c)(c-a)}=\frac{b(a-b)(a-c)-c(a-c)(a-b)}{(a-b)(b-c)(c-a)}=\frac{(a-b)(b-c)(a-c)}{(a-b)(b-c)(c-a)}=-1$Ta có $\frac{a^2}{(b-c)^2}+\frac{b^2}{(c-a)^2}+\frac{c^2}{(a-b)^2}=(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})^2-2P\geq -2P=2$
cái này là bđt $ x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)$