Inequality with $ab+bc+ca=3$

Question

Let $a$, $b$, $c>0$ and $ab+bc+ca=3$, prove that $\sum_{cyc}{\frac{a^2+b^2}{a+b+1}}\geq \frac{6abc+6}{a+b+c+3}$.

I have tried using Holder $\sum_{cyc}{\frac{a^2+b^2}{a+b+1}}\geq \frac{2(a+b+c)^2}{\sum_{cyc}{a+b+1}}$

Answer 1

Your idea works!

Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v>0$ and $abc=w^3$.

Hence, $u\geq v\geq w$ and by C-S we obtain: $$\sum_{cyc}\frac{a^2+b^2}{a+b+1}\geq\frac{2(a+b+c)^2}{\sum\limits_{cyc}(2a+1)}=\frac{2(a+b+c)^2}{2(a+b+c)+3}=$$ $$=\frac{18u^2}{6u+3v}=\frac{6u^2}{2u+v}.$$ Thus, it's enough to prove that $$\frac{6u^2}{2u+v}\geq\frac{6(abc+1)}{a+b+c+3}$$ and since $w^3\leq v^3$, it's remains to prove that $$\frac{6u^2}{2u+v}\geq\frac{6(v^3+v^3)}{(3u+3v)v}$$ or $$\frac{u^2}{2u+v}\geq\frac{2v^2}{3(u+v)}$$ or $$(u-v)(3u^2+6uv+2v^2)\geq0.$$ Done!