<s>[INST] 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.
 [/INST] Given that $a+b+c=6$ we can substitute $a=6-b-c$ into $A$ to get
$$A=(6-b-c)^2(bc+1)+2(b^2+c^2),$$
where $b,c>0$ and $b+c<6$. Differentiating with respect to $b$ and $c$
yields
$$\frac{\partial A}{\partial b}=-2(6-b-c)(bc+1)+(6-b-c)^2c+4b,$$
$$\frac{\partial A}{\partial b}=-2(6-b-c)(bc+1)+(6-b-c)^2b+4c.$$
If there is a maximum then both derivatives must equal 0 in such a point, and hence 
$$0=\frac{\partial A}{\partial b}-\frac{\partial A}{\partial c}=(c-b)((6-b-c)^2-4),$$
so either $b=c$ or $b+c=4$. In the latter case we get $c=4-b$ and so
$$A=2(bc+1)+2(b^2+c^2)=2(b(4-b)+1)+2(b^2+(4-b)^2)=2b^2-8b+34,$$
which does not assume a maximum on $(0,4)$. In the former case where $b=c$ we get
$$A=(6-2b)(b^2+1)+4b^2=-2b^3+10b^2-2b+6,$$
which does not assume a maximum on $(0,3)$. So the maximum does not exist.
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