<s>[INST] maximum value of expression $6bc+6abc+2ab+3ac$ If $a,b,c>0$ and $a+2b+3c=15,$ then finding maximum value of $6bc+6abc+2ab+3ac$ is
with the help of AM - GM inequality 
$4ab\leq (a+b)^2$ and $4bc \leq (b+c)^2$ and $\displaystyle 4ca \leq (c+a)^2$ 
and $27(abc)\leq (a+b+c)^3$
could some help me, thanks
 [/INST] Let $a=5x$, $2b=5y$ and $3c=5z$.
Hence, $x+y+z=3$ and by AM-GM $bc+6abc+2ab+3ac=$
$$=25(xy+xz+yz)+ 125xyz\leq 25 \frac{(x+y+z)^2}{3}  + 125 \left(\frac{x+y+z}{3}\right)^3=200.$$
The equality occurs for $x=y=z=1$, which says that the answer is $200$.
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