Prove: $\sum\limits_{cyc}\frac{1}{b^2+bc+c^2} \ge \frac{2}{ab+bc+ca}+\frac{a+b+c}{3(a^3+b^3+c^3)}.$ with $a,b,c \ge 0 : ab+bc+ca>0.$

Question

Let $a,b,c \ge 0 : ab+bc+ca >0.$ Prove that $$\dfrac{1}{a^2+ab+b^2}+\dfrac{1}{b^2+bc+c^2}+\dfrac{1}{c^2+ca+a^2} \ge \dfrac{2}{ab+bc+ca}+\dfrac{a+b+c}{3(a^3+b^3+c^3)}.$$

The big problem here is inequality holds iff $a=b=c$ and $a=b, c=0$ and permutations.

I tried Cauchy - Schwarz: $$\sum\limits_{cyc}\dfrac{1}{b^2+bc+c^2} \ge \dfrac{(a+b+c)^2}{\sum\limits_{cyc}a^2(b^2+bc+c^2)}.$$ But the following is not nice. I also tried full expanding and checked it by Maple, It's true. But I don't have any nice solution for it.

Please help me. Thank you very much.

Answer 1

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that: $$\frac{\sum\limits_{cyc}(a^2+ab+b^2)(a^2+ac+c^2)}{\prod\limits_{cyc}(a^2+ab+b^2)}\geq\frac{2}{ab+ac+bc}+\frac{a+b+c}{3(a^3+b^3+c^3)}$$ or $$\frac{27(3u^4-3u^2v^2+v^4)}{27(3u^2v^4-u^3w^3-v^6)}\geq\frac{2}{3v^2}+\frac{u}{27u^3-27uv^2+3w^3}$$ or $f(w^3)\geq0,$ where $$f(w^3)=\frac{3u^4-3u^2v^2+v^4}{3u^2v^4-u^3w^3-v^6}-\frac{2}{3v^2}-\frac{u}{27u^3-27uv^2+3w^3}.$$ But $$f'(w^3)=\frac{u^3(3u^4-3u^2v^2+v^4)}{(3u^2v^4-u^3w^3-v^6)^2}+\frac{3u}{(27u^3-27uv^2+3w^3)^2}>0,$$ which says that $f$ increases(we can see it also without derivative) and it's enough to prove our inequality for a minimal value of $w^3$, which by $uvw$ happens in the following cases.

1. $w^3=0.$

Let $c=0$ and $b=1$.

We obtain: $$(a-1)^2(3a^4+5a^2+3)\geq0.$$

2. Two variables are equal.

Let $b=c=1$.

We obtain: $$(a-1)^2a(2a^3+a^2+7a+11)\geq0$$ and we are done!

About $uvw$ see here: https://artofproblemsolving.com/community/c6h278791