If $a+b+c=3$, find the greatest value of $a^2b^3c^2$.

If $a+b+c=3$, and $a,b,c>0$ find the greatest value of $a^2b^3c^2$.

I have no idea as to how I can solve this question. I only require a small hint to start this question. It would be great if someone could help me with this.

Questions like these are about tricks. Here's one you should remember.

Rewrite $a+b+c = 3$ as $2\frac{a}{2} + 3\frac b3 + 2\frac c2 =3$. Use the AM-GM inequality: $$ \frac{2\frac{a}{2} + 3\frac b3 + 2\frac c2}{7} \geq \sqrt[7]{\frac{a^2b^3c^2}{4 \cdot 27 \cdot 4}} $$ So we simplify: $$ a^2b^3c^2 \leq 432 \left(\frac 37\right) ^7 = \frac{944784}{823543} \simeq 1.1472 $$ Equality is attained (and that's important!) When all the terms are equal i.e. $\frac{a}{2} = \frac{b}{3} = \frac{c}{2}$. You can check this happens when $a = \frac 67, b = \frac 97, c= \frac 67$.

When $a,b,c$ are as above, $a^2b^3c^2 = \frac{6^4 9^3}{7^7} = \frac{944784}{823543}$.