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.

Next to the usage of Lagrange multipliers, one can also solve it directly function-wise by assuming $c(a,b)=6-a-b$, which leads to the following derivatives to find the extrema:
$$\left\{\begin{align}
\frac{\partial A}{\partial a} &= (ab-2)(12-3a-2b)&=0 \\
\frac{\partial A}{\partial b} &= (a^2-4)(6-a-2b)&=0 \\
\end{align}\right.$$
This gives us various solutions:


*

*${\partial A}/{\partial b}=0$ gives us $a=2$ leading to:


*

*$(a,b,c)=(2,1,3)$ with A = 36

*$(a,b,c)=(2,3,1)$ with A = 36


*${\partial A}/{\partial b}=0$ gives us $2b=6-a$ resembling $b=c$, leading to:


*

*$6-2a=0$ or $(a,b,c)=(3,3/2,3/2)$ with A = 38.25

*$ab=2$ which creates a quadratic eqution having : $(a,b,c)=(3\pm\sqrt5,(3\mp\sqrt 5)/2,(3\mp\sqrt 5)/2)$ with A = 32
From all these five points, the Hessian matrix shows that they are saddle points, with two exceptions, $(a,b,c)=(3-\sqrt5,(3+\sqrt 5)/2,(3+\sqrt 5)/2)$ which is a local minimum and $(a,b,c)=(3,3/2,3/2)$ which is a local maximum.
As the domain of investigation is $a>0,b>0,c>0$ we should also investigate these various planes for extrema:


*

*$a=0$ gives that $A=72-24b +4b^2$ which has a local minimum in $b=3$ with A = 36

*$b=0$ gives that $A=72-24a+3a^2$ which has a local minimum in $a=4$ with A = 24
Note: these points are not true extrema of the function $A(a,b,c(a,b))$
The real maximal value, however, is reached when $(a,b,c)$ tends to $(0,0,6)$ with A=72.
Note: this point is the highest value reached, which lays just outside of the domain and is a supremum.
