Ta có các bất đẳng thức sau: $3=a+b+c\geq 3\sqrt[3]{abc} \Rightarrow abc\leq 1$$\Leftrightarrow 9={{(a+b+c)}^{2}}\ge 3(ab+bc+ca)\Rightarrow ab+bc+ca\le 3$$3\sqrt[3]{{{a}^{2}}{{b}^{2}}}(3\sqrt[3]{{{c}^{2}}}+1)=9\sqrt[3]{{{a}^{2}}{{b}^{2}}{{c}^{2}}}+3\sqrt[3]{{{a}^{2}}{{b}^{2}}}\le [3({{a}^{2}}{{b}^{2}}{{c}^{2}}+1+1)]+({{a}^{2}}{{b}^{2}}+1+1)$$(AM-GM)$

$\Rightarrow    3\sqrt[3]{a^{2}b^{2}}(3\sqrt[3]{c^{2}}+1)-8 \leq 3a^{2}b^{2}c^{2}+a^{2}b^{2}$$\Rightarrow \dfrac{a^{2}+1}{3\sqrt[3]{a^{2}b^{2}}(3\sqrt[3]{c^{2}}+1)-8} \geq \dfrac{a^{2}+1}{3a^{2}b^{2}c^{2}+a^{2}b^{2}}$

$\Rightarrow \sum_{cyc}\dfrac{a^{2}+1}{3\sqrt[3]{a^{2}b^{2}}(3\sqrt[3]{c^{2}}+1)-8} \geq \sum_{cyc}\dfrac{a^{2}+1}         {3a^{2}b^{2}c^{2}+a^{2}b^{2}}$

Mà$ \sum_{cyc}\dfrac{a^{2}+1}{3a^{2}b^{2}c^{2}+a^{2}b^{2}} \geq \sum_{cyc}\dfrac{(a+1)^{2}}{6a^{2}b^{2}c^{2}+2a^{2}b^{2}} \geq$$\dfrac{(a+b+c+1+1+1)^{2}}{18a^{2}b^{2}c^{2}+2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})}= \dfrac{18}{   9a^{2}b^{2}c^{2}+a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}}$  $(1)$

Mặt khác:$9a^{2}b^{2}c^{2}+a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2} \leq 9abc.1 + a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}$

              $= 2abc(a+b+c)+ a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2} + 3abc = (ab+bc+ca)^{2}+3abc \leq 3^{2}+3= 12$

             $\Rightarrow 9a^{2}b^{2}c^{2}+a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2} \leq 12$  $(2)$

Từ $(1)$ và $(2)$ thì $\sum_{cyc}\dfrac{a^{2}+1}{3a^{2}b^{2}c^{2}+a^{2}b^{2}} \geq \dfrac{3}{2}$$\Rightarrow \sum_{cyc}\dfrac{a^{2}+1}{3\sqrt[3]{a^{2}b^{2}}(3\sqrt[3]{c^{2}}+1)-8} \geq \dfrac{3}{2}$