4
$\begingroup$

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.

$\endgroup$
  • 5
    $\begingroup$ 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
  • $\begingroup$ 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. $\endgroup$ – астон вілла олоф мэллбэрг Nov 24 '16 at 5:30
  • 1
    $\begingroup$ @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! $\endgroup$ – астон вілла олоф мэллбэрг Nov 24 '16 at 5:31
14
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.