Question

If a, b, c are positive real number such that a = b+ c = 18, the find the maxiumum valuye $a^2{b}^3{C}^4$

Answer 1

Applying $A.M⩾G.M$ for $\frac{a}{2},\frac{a}{2},\frac{b}{3},\frac{b}{3},\frac{b}{3},\frac{c}{4},\frac{c}{4},\frac{c}{4},\frac{c}{4}$

(to get $a^2{b}^3{c}^4$ on R.H.S)

$\Rightarrow \frac{a+b+c}{9}⩾(\frac{a^2}{4}\times \frac{b^3}{27}\times \frac{c^4}{256}{)}^{1/9}$

$\Rightarrow (\frac{18}{9}{)}^9\times (4\times 27\times 256)⩾a^2{b}^3{c}^4$

$\Rightarrow a^2{b}^3{c}^4⩽2^3\left(27648\right)$

$\Rightarrow 2^9\times 27648$ is greatest value of $a^2{b}^3{c}^4$