$(a b+b c+a c)^2\left[\frac{1}{(a-b)^4}+\frac{1}{(b-c)^4}+\frac{1}{(c-a)^4}\right] \geq \frac{33}{16}$

Question

Let a,b,c be distinct real number such that $a+b+c=0$ .Prove that the following inequality holds

$$(a b+b c+a c)^2\left[\frac{1}{(a-b)^4}+\frac{1}{(b-c)^4}+\frac{1}{(c-a)^4}\right] \geq \frac{33}{16}$$

I have tried to find some related problems like the one in VMO 2008, but I still can't solve this problem since the condition of a,b,c is real number. Also, I also tried to use mixing variables because equality occurs when a=-b,c=0 but doesn't give result.

Please help me

Answer 1

Since $\prod\limits_{cyc}(ab)=a^2b^2c^2\geq0,$ we can assume that $ab\geq0$ and since $c=-a-b,$ we need to prove that: $$a^2b^2(a+b)^2(156a^6+456a^5b+615a^4b^2+470a^3b^3+615a^2b^4+456ab^5+152b^6)\geq0,$$ which is obvious.

Answer 2

Another way.

Let $b=a+u$ and $c=a+v$, where $uv\leq0.$

Thus, $a=-\frac{u+v}{3}$ and by AM-GM we obtain: $$(ab+ac+bc)^2=(3a^2+2(u+v)a+uv)=\left(\frac{(u+v)^2}{3}-\frac{2(u+v)^2}{3}+uv\right)^2=$$ $$=\frac{1}{9}(u^2-uv+v^2)^2\geq\frac{1}{9}(-2uv-uv)^2=u^2v^2.$$ Thus, it's enough to prove that: $$u^2v^2\left(\frac{1}{u^4}+\frac{1}{v^4}+\frac{1}{(u-v)^4}\right)\geq\frac{33}{16}$$ or $$\frac{u^2}{v^2}+\frac{v^2}{u^2}-2\geq\frac{1}{16}-\frac{u^2v^2}{(u-v)^4}$$ or $$\frac{(u^2-v^2)^2}{u^2v^2}\geq\frac{(u+v)^2(u^2-6uv+v^2)}{16(u-v)^4}$$ or $$16(u-v)^6\geq(u^2-6uv+v^2)u^2v^2.$$ But by AM-GM again $$u^2-6uv+v^2\leq2(u-v)^2$$ and it's enough to prove that: $$8(u-v)^4\geq u^2v^2,$$ which is true by AM-GM again: $$8(u-v)^4=8(u^2+v^2-2uv)^2\geq8(-2uv-2uv)^2\geq128u^2v^2\geq u^2v^2.$$