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. $\frac{3^4{2}^{10}}{7^7}$
• B. $\frac{3^{10}{2}^4}{7^7}$
• C. $\frac{3^2{2}^{12}}{7^7}$
• D. $\frac{3^{12}{2}^2}{7^7}$

Answer 1

Answer: (B)

$\frac{3^{10}{2}^4}{7^7}$

Given $a+b+c=3$, then $a^2{b}^3{c}^2$ (greater value)

$a+b+c=\frac{a}{2}+\frac{a}{2}+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}+\frac{c}{2}+\frac{c}{2}=3$

Now we $A.M\ge G.M$

So, $\frac{\frac{a}{2}+\frac{a}{2}+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}+\frac{c}{2}+\frac{c}{2}}{7}\ge {[\frac{a}{2}\times \frac{a}{2}\times \frac{b}{3}\times \frac{b}{3}\times \frac{b}{3}\times \frac{c}{2}\times \frac{c}{2}]}^{1/7}$

$\therefore \frac{3}{7}\ge {(\frac{a^2}{4}\times \frac{b^3}{27}\times \frac{c^2}{4})}^{1/7}$

$\Rightarrow {\left(\frac{3}{7}\right)}^7\times 2^2\times 3^3\times 2^2\ge a^2{b}^3{c}^2$

So greatest value is $\frac{3^{10}{2}^4}{7^7}$