Inequality by Kristian Vasilev

$Let\quad B=6\left( \frac { a }{ c } +\frac { b }{ a } +\frac { c }{ b } \right) \quad and\quad C=\frac { { a }^{ 2 }+9 }{ c } +\frac { { b }^{ 2 }+9 }{ a } +\frac { { c }^{ 2 }+9 }{ b } \\ \Rightarrow A=B+C\\ Using\quad the\quad inequality\quad \left( \frac { a }{ c } +\frac { b }{ a } +\frac { c }{ b } \right) \ge 3\quad we\quad get\quad B\ge 18\\ Using\quad the\quad inequality\quad { x }^{ 2 }+{ y }^{ 2 }\ge 2xy\quad we\quad get\quad C\ge 6\left( \frac { a }{ c } +\frac { b }{ a } +\frac { c }{ b } \right) =B\Rightarrow C\ge 18\\ \Rightarrow A\ge 36$

By the Cauchy-Schwarz Inequality, we have: $\Sigma \frac{(a+3)^2}{c} \geq \frac{(9+a+b+c)^2}{a+b+c}$ Now by the AM-GM inequality: $(9+a+b+c)^2 \geq 36(a+b+c)$ This gives us a minimum of $\boxed{36}$ which is achieved when $a,b,c=3$ .

By Cauchy-Schwarz's Inequality, we have $\Sigma \frac{(a+3)^2}{c} \ge \frac{(a+b+c+9)^2}{a+b+c}$ We have that $(a+b+c+9)^2=(a+b+c)^2+18(a+b+c)+81$ . Thus, we want to minimize $\frac{(a+b+c)^2+18(a+b+c)+81}{a+b+c}$ $=(a+b+c)+\frac{81}{a+b+c}+18$ By AM-GM, we have $(a+b+c)+\frac{81}{a+b+c} \ge 18$ Thus, the minimum value of $\boxed{36}$ , when $a,b,c=3$