Let: $a,b,c \in \mathbb{R}$ and $a\neq b\neq c$. Prove that with $k \in [-1;2]$ we have:
$(a^2+b^2+c^2+k(ab+bc+ca))(\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2})\geq \frac{9}{4}(2-k)$
By expanding by computer, (because I don't have any idea :(( ): $$\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}=\frac{(a^2+b^2+c^2-ab-bc-ca)^2}{(a-b)^2(b-c)^2(c-a)^2}$$
So the inequality can be rewritten as:
$$(a^2+b^2+c^2+k(ab+bc+ca))(\frac{(a^2+b^2+c^2-ab-bc-ca)^2}{(a-b)^2(b-c)^2(c-a)^2})\geq \frac{9}{4}(2-k)$$
But I can't solve it from here :( Can anyone help me? Thanks for your help!!
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that:$$(9u^2-6v^2+3kv^2)\sum_{cyc}(a-b)^2(a-c)^2-\frac{9}{4}(2-k)\cdot27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)\geq0,$$ which says that a coefficient before $w^6$ is equal to $\frac{243(2-k)}{4},$ which says that the inequality $$(9u^2-6v^2+3kv^2)\sum_{cyc}(a-b)^2(a-c)^2-\frac{9}{4}(2-k)\cdot27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)\geq\frac{2-k}{3}\left(\sum_{cyc}\left(a^3-\frac{3}{2}(a^2b+a^2c)+2abc\right)\right)^2$$ is a linear inequality of $w^3$, which by $uvw$ says that it's enough to prove the last inequality for equality case of two variables, which for $c=b$ gives $$(k+1)(a-b)^4(a+2b)^2\geq0$$ and we are done.