<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] We are nearly done, but we need something other than $a,b,c$ because of the condition of the problem. 
By Cauchy-Schwarz we have $$3(x^2+y^2+z^2) \ge (x+y+z)^2 \iff (x+y+z)^2 \ge 3(xy+yz+zx)$$
So put $x=a, y=2b, z=3c$ to get 
$$225=(a+2b+3c)^2 \ge 3(2ab+3ac+6bc) \iff 75 \ge 2ab+3ac+6bc  $$
Also, note that by AM-GM $$15=a+2b+3c \ge 3 \sqrt[3]{6abc} \iff 125 \ge 6abc $$
Note that both inequalities have equalities when $a=2b=3c$.
So the answer is $200$. 
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