Find the maximum value of $A$ Let $a;b;c>0$ such that $a+b+c=6$. Find the maximum value of  $A=a^2bc+a^2+2b^2+2c^2$

WLOG $b\ge c$. I see maximum value of $A=36$ at $(a;b;c)=(2;1;3)$
So i need to prove $A\le 36$. Or I will prove 
$(a+b+c)^4\ge 36a^2bc+(a^2+2b^2+2c^2)(a+b+c)^2$
Or $(2a-b-c)(b^3+c^3+a^2b+a^2c+2ab^2+2ac^2-12abc-b^2c-bc^2)\ge 0$
Then Im stuck here, help me solve it.

With the help of the Lagrange multipliers and $f(a,b,c) =  a^2 b c + a^2 + 2 b^2 + 2 c^2$
$$
L(a,b,c,\lambda) =f(a,b,c)+\lambda(a+b+c-6)
$$
the stationary points are the solutions for
$$
\nabla L = \left\{
\begin{array}{c}
 2 b c a+2 a+\lambda =0 \\
 c a^2+4 b+\lambda =0 \\
 b a^2+4 c+\lambda =0 \\
 a+b+c-6=0 \\
\end{array}
\right.
$$
giving
$$
\left[
\begin{array}{ccccc}
a & b & c & \lambda & f(a,b,c)\\
 -2 & -1 & 9 & -32 & \color{red}{132} \\
 -2 & 9 & -1 & -32 & \color{red}{132} \\
 2 & 1 & 3 & -16 & \color{green}{36} \\
 2 & 3 & 1 & -16 & \color{green}{36} \\
 3 & \frac{3}{2} & \frac{3}{2} & -\frac{39}{2} & \color{green}{\frac{153}{4}} \\
 3-\sqrt{5} & \frac{1}{2} \left(3+\sqrt{5}\right) & \frac{1}{2} \left(3+\sqrt{5}\right) & -12 & \color{green}{32} \\
 3+\sqrt{5} & \frac{1}{2} \left(3-\sqrt{5}\right) & \frac{1}{2} \left(3-\sqrt{5}\right) & -12 & \color{green}{32} \\
\end{array}
\right]
$$
In red non feasible solutions and in green feasible solutions. The maximum corresponds to $\frac{153}{4}= 38.25$
NOTE
The same solution can be obtained under the tangency condition between $f(a,b,c)-\mu=0$ and $a+b+c-6=0$ by calculating the values for $\mu$
