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

You can proceed as follows:
$$a+2b+3c=15\tag1$$
From $(1)$ , we have by AM-GM inequality,
$$\frac{a+2b+3c}{3} \ge \sqrt[3]{6abc}$$
$$\implies \frac{15}{3} \ge \sqrt[3]{6abc}$$
$$\implies 5^3 \ge 6abc$$
Again from $(1)$ , we have by AM-GM inequality,
$$\frac{a+2b+3c}{2} \ge \frac{a+2b}{2} \ge \sqrt{2ab}$$
$$\frac{15}{2} \ge \sqrt{2ab}$$
$$\frac{225}{4} \ge 2ab$$
Again from $(1)$ , we have by AM-GM inequality,
$$\frac{a+2b+3c}{2} \ge \frac{2b+3c}{2} \ge \sqrt{6bc}$$
$$\frac{15}{2} \ge \sqrt{6bc}$$
And so on .
Hope this helps you and you can complete the answer.
