Question

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

Answer 1

$a+b+c=3$,

we need $a^2.b^3.c^2$,

so, $a+b+c=3$,

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

Now we use AM-GM,

$\frac{\frac{a}{2}+\frac{a}{2}+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}+\frac{c}{2}+\frac{c}{2}}{7}={\left(\frac{a^2}{2^2}\times \frac{b^3}{3^3}\times \frac{c^2}{2^2}\right)}^{\frac{1}{7}}$

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

$\frac{3^{10}\times 2^4}{7^7}\ge a^2{b}^3{c}^2$,