Let $a,b,c>0$ such that $ab+bc+ca=3$. Prove that $$a^2+b^2+c^2+3abc\ge \sqrt{3\left(\frac{1}{\left(a+b\right)^2}+\frac{1}{\left(b+c\right)^2}+\frac{1}{\left(a+c\right)^2}\right)}+3abc\left(\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca}\right)$$
We have: $\frac{1}{3-ab}\le \frac{ab}{4}+\frac{1}{4}$$\Leftrightarrow -\frac{\left(ab-1\right)^2}{4\left(ab-3\right)}\le 0\forall 0<ab\le 3$
Then I used SOS but $\sqrt{3\left(\frac{1}{\left(a+b\right)^2}+\frac{1}{\left(b+c\right)^2}+\frac{1}{\left(a+c\right)^2}\right)}$ is hard to used
The hint.
Prove that $$a^2+b^2+c^2\geq\frac{1}{2}\sum_(a^2b+a^2c)$$ and $$\sqrt{3\sum_{cyc}(a+b)^2(a+c)^2}\leq2\sum_{cyc}(a^2+ab).$$ Thus, it's enough to prove that $$\frac{1}{2}\sum_{cyc}(a^2b+a^2c)+3abc\geq\frac{2(ab+ac+bc)^2\sum\limits_{cyc}(a^2+ab)}{9(a+b)(a+c)(b+c)}+(ab+ac+bc)\sum_{cyc}\frac{ab}{a+b},$$ which is smooth.