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.

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    $\begingroup$ Rewrite $a+b+c=3$ as $a/2+a/2+b/3+b/3+b/3+c/2+c/2=3$ and use the property Arithmetic Mean $\geq$ Geometric Mean. $\endgroup$ – vighnesh_9 Nov 24 '16 at 5:27
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    $\begingroup$ Are $a,b,c$ assumed to be positive? $\endgroup$ – dxiv Nov 24 '16 at 5:28
  • $\begingroup$ @dxiv I think so ,right? Otherwise, just look at $a=-n,c=-n,b=2n+3$, then for positive $n$ it would evaluate to $n^4(2n+3)^3$, which is increasing with $n$ for example. $\endgroup$ – Teresa Lisbon Nov 24 '16 at 5:30
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    $\begingroup$ @астонвіллаолофмэллбэрг Right, of course. But such assumptions need to be stated in the question, not left to second-guess. $\endgroup$ – dxiv Nov 24 '16 at 5:31
  • $\begingroup$ @dxiv Yes, that is correct, though! $\endgroup$ – Teresa Lisbon Nov 24 '16 at 5:31

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

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