<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] For $(a,b,c)\rightarrow(0,6,0)$ we have $A\rightarrow72$ and since
$$72-a^2bc-a^2-2b^2-2c^2\geq0$$ it's
$$2(a+b+c)^4-(a+b+c)^2(a^2+2b^2+2c^2)-36a^2bc\geq0$$ or
$$a^4+6a^3b+6a^3c+9a^2b^2+9a^2c^2-14a^2bc+4ab^3+4ac^3+20b^2ac+20c^2ab+4ab^3+4ac^3+8b^2c^2\geq0,$$
which is obvious, we see that the maximum does not exist, but the supremum is $72$.
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