By $AM-GM$ we have $$(3a^2+1)^2+2(1+\frac{3}{b})^2=(3a^2+1)^2+(1+\frac{3}{b})^2+(1+\frac{3}{b})^2 \geq 3\sqrt[3]{(3a^2+1)^2(1+\frac{3}{b})^2(1+\frac{3}{b})^2}$$ and $1+3x=1+x+x+x \geq \sqrt[4]{x^3}$ thus for $x=a^2$ and $x=\frac{1}{b}$ we get $(1+3a^2)^2 \geq 16a^3$ and $(1+\frac{3}{b})^4 \geq 16^2 \frac{1}{b^3}$ thus $$(3a^2+1)^2+2(1+\frac{3}{b})^2 \geq 3.16 \sqrt[3]{\frac{a}{b}}=48\sqrt[3]{\frac{a}{b}}$$ and we do the same with the other terms. Hence $$(3a^2+1)^2+2(1+\frac{3}{b})^2][(3b^2+1)^2+2(1+\frac{3}{c})^2][(3c^2+1)^2+2(1+\frac{3}{a})^2]\geq 48^3\sqrt[3]{\frac{a}{b}\frac{b}{c}\frac{c}{a}}=48^3$$

QMHM & AMGM

\[[(3a^2+1)^2+2(1+\frac{3}{b})^2][(3b^2+1)^2+2(1+\frac{3}{c})^2][(3c^2+1)^2+2(1+\frac{3}{a})^2]\geq48^3\] $(3a^2+1)^2+2(1+\frac{3}{b})^2=(3a^2+1)^2+(1+\frac{3}{b})^2+(1+\frac{3}{b})^2$ By C.S $\Longrightarrow$ $(3a^2+1)^2+2(1+\frac{3}{b})^2=(3a^2+1)^2+(1+\frac{3}{b})^2+(1+\frac{3}{b})^2\geq \frac{(3a^2+\frac{6}{b}+3)^2}{3}$ then BY AM-GM $3a^2+3\geq 6a$ Then equality takes the form $\frac{(6a+\frac{6}{b})^2}{3}$ then BY AM-GM $6a+\frac{6}{b}\geq12\sqrt{\frac{a}{b}}$ $\Longrightarrow$ $\frac{(6a+\frac{6}{b})^2}{3}\geq48\frac{a}{b}$ Then $48\frac{a}{b}48\frac{b}{c}48\frac{c}{a}=48^3$ Anar Abbas

my solution= by cauchy we have $(3a^2+1)^2+2(1+3b)^2\ge \frac{(3a^2+\frac{6}{b}+3)^2}{3}$ and by AM-GM we have $\frac{(3a^2+\frac{6}{b}+3)^2}{3}\ge \frac{144\sqrt a}{3\sqrt b}$ as $3a^2+\frac{6}{b}+3\ge 12\sqrt {\frac{a}{b}} $ so combining we get \[[(3a^2+1)^2+2(1+\frac{3}{b})^2][(3b^2+1)^2+2(1+\frac{3}{c})^2][(3c^2+1)^2+2(1+\frac{3}{a})^2]\geq48^3\] so we are done

By AM-GM $\prod_{cyc} \left((3a^2 + 1)^2 + 2\left(1 + \dfrac{3}{b}\right)^2\right) \ge 48^3$ $\prod_{cyc} \left((3a^2 + 1)^2 + \left(1 + \dfrac{3}{b}\right)^2+ \left(1 + \dfrac{3}{b}\right)^2\right) \ge 48^3$ $\prod_{cyc} \left((3a^2 + 1)^{\frac{2}{3}}(1 + \frac{3}{b})^{\frac{4}{3}}\right) \ge 16^3$ $\prod_{cyc} \left((3a^2 + 1)^{\frac{1}{3}}(1 + \frac{3}{a})^{\frac{2}{3}}\right) \ge 4^3$ Which is just Holder.

Just a simplification of @aditya21 's solution. a^2 + 2/b +1 = a^2 + 1/b + 1/b +1 >= 4 sqrt (a/b) from AM-GM

My Solution: So its basically given that \[\prod_{cyc}\left((3a^2+1)^2+2\left(1+\frac3b\right)^2\right)\geq 48^3\]by AM-GM we can write this \[1+a^2+a^2+a^2\geq 4a\sqrt{a}\]and we also know that\[1+\frac1b+\frac1b+\frac1b\geq 4\sqrt[4]{\frac{1}{b^3}}\]by this we get that \[\prod_{cyc}\left((3a^2+1)^2+2\left(1+\frac3b\right)^2\right)\geq \prod_{cyc}\left(16a^3+16\sqrt{\frac{1}{b^3}}+16\sqrt{\frac{1}{b^3}}\right)\geq 48^3\]Applying AM-GM again we have\[\prod_{cyc}\left(3\sqrt[3]{16^3\frac{a^3}{\sqrt{b^6}}}\right)=\prod_{cyc}\left(16\cdot3\frac{a}{b}\right)=\left(\frac{48a}{b}\right)\left(\frac{48b}{c}\right)\left(\frac{48c}{a}\right)\geq 48^3\]which we can clearly see that this is true !$\blacksquare$

Nice! Here is an elementary Solution only involving $AM$ $GM$ (Which took me only an eternity to write )