Question

If $a+b+c=3$ and $a>0,b>0,c>0$, then the greatest value of $a^2{b}^3{c}^2$ is

• A. $\frac{3^{10}\cdot 2^4}{7^7}$
• B. $\frac{3^{10}\cdot 2^7}{7^7}$
• C. $\frac{3^7\cdot 2^4}{7^7}$
• D. $\frac{3^7\cdot 2^4}{7^4}$

Answer 1

The correct option is A $\frac{3^{10}\cdot 2^4}{7^7}$

Taking $A.M$ and $G.M$ of $7$ numbers $\frac{a}{2},\frac{a}{2},\frac{b}{3},\frac{b}{3},\frac{b}{3},\frac{c}{2},\frac{c}{2}$, we get

$\frac{2\cdot \frac{a}{2}+3\cdot \frac{b}{3}+2\cdot \frac{c}{2}}{7}\ge {\left\{{\left(\frac{a}{2}\right)}^2{\left(\frac{b}{3}\right)}^3{\left(\frac{c}{2}\right)}^2\right\}}^{\frac{1}{7}}$
$\Rightarrow \frac{3}{7}\ge {\left(\frac{a^2{b}^3{c}^2}{\left(2^2\right)\left(3^3\right)\left(2^2\right)}\right)}^{\frac{1}{7}}$
$\Rightarrow \frac{3^7}{7^7}\ge \frac{a^2{b}^3{c}^2}{\left(2^2\right)\left(3^3\right)\left(2^2\right)}$
$\Rightarrow a^2{b}^3{c}^2\le \frac{3^{10}\cdot 2^4}{7^7}$
$\therefore$ Greatest value of $a^2{b}^3{c}^2=\frac{3^{10}\cdot 2^4}{7^7}$