Question

If $2a+b+3c=1$ and a, b, c $>0$, then find the maximum value of $a^4{b}^2{c}^2$

Answer 1

Consider $\frac{a}{2},\frac{a}{2},\frac{a}{2},\frac{a}{2},\frac{b}{2},\frac{b}{2},\frac{3c}{2},\frac{3c}{2}$

A.M$\ge$G.M

$\frac{2a+b+3c}{8}\ge {(\frac{a^4}{16}\times \frac{b^2}{4}\times \frac{9c^2}{4})}^{\frac{1}{8}}$

$\Rightarrow \frac{9a^4{b}^2{c}^2}{256}\le {\left(\frac{1}{8}\right)}^8$

$\Rightarrow a^4{b}^2{c}^2\le {\left(\frac{1}{8}\right)}^8\times \frac{256}{9}$

$\Rightarrow a^4{b}^2{c}^2\le \frac{1}{{\left(2^3\right)}^8}\times \frac{2^8}{9}$

$\Rightarrow a^4{b}^2{c}^2\le \frac{1}{2^{24}}\times \frac{2^8}{9}$

$\Rightarrow a^4{b}^2{c}^2\le \frac{1}{2^{24-8}}\times \frac{1}{9}$

$\therefore a^4{b}^2{c}^2\le \frac{1}{9\times 2^{16}}$