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

  • $\begingroup$ Can you understand? $\endgroup$
    – S.C.B.
    Feb 2, 2017 at 15:05

3 Answers 3


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 $200$.


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 $200$.


You can proceed as follows:


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.


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