Anyone??

Observe that \begin{align*} \sqrt{a+3b+\frac{2}{c}} \cdot \sqrt{b+3c+\frac{2}{a} } &=\sqrt{ \left( a+b+b+b+\frac{2}{c} \right) \left( \frac{1}{a}+\frac{1}{a}+b+c+2c \right) } \\ & \ge 1+\sqrt{ \frac{b}{a} }+b+ \sqrt{bc}+2 \\ &= 3+b+\sqrt{ \frac{b}{a} }+\sqrt{bc} \end{align*}Therefore, \begin{align*} \left( \sum_{cyc} \sqrt{a+3b+\frac{2}{c}} \right) ^2 &= \sum_{cyc} \left( a+3b+\frac{2}{c} \right) + 2 \sum_{cyc} \sqrt{a+3b+\frac{2}{c}} \cdot \sqrt{b+3c+\frac{2}{a} } \\ & \ge \sum_{cyc} \left( a+3b+\frac{2}{c} \right) + 2 \sum_{cyc} \left( 3+b+\sqrt{ \frac{b}{a} }+\sqrt{bc} \right) \\ &= 18+ 6\sum_{cyc} a +2\sum_{cyc} \sqrt{ \frac{b}{a} } +2\sum_{cyc} \left( \frac{1}{a}+\sqrt{bc} \right) \\ & \ge 18+18+6+12=54. \end{align*}

Rickyminer Very nice solution

jasperE3 wrote: Show that for all positive reals $a,b,c$ with $a+b+c=3$, $$\sum_{\text{cyc}}\sqrt{a+3b+\frac2c}\ge3\sqrt6.$$ https://artofproblemsolving.com/community/c6h1748538p11390553