Question

If $x+y+z=1$, $x,y,z>0$. Then greatest value of $x^2{y}^3{z}^4$ is

• A. $\frac{2^9}{3^5}$
• B. $\frac{2^{10}}{3^{15}}$
• C. $\frac{2^{15}}{3^{10}}$
• D. $\frac{2^{10}}{3^{10}}$

Answer 1

The correct option is B $\frac{2^{10}}{3^{15}}$

$x+y+z=1\Rightarrow 2\cdot \frac{x}{2}+3\cdot \frac{y}{3}+4\cdot \frac{z}{4}=1$
Using A.M., G.M. inequality for the terms $\frac{x}{2},\frac{x}{2},\frac{y}{2},\frac{y}{2},\frac{y}{2},\frac{z}{2},\frac{z}{2},\frac{z}{2},\frac{z}{2}$ $\frac{2\cdot \left(\frac{x}{2}\right)+3\cdot \left(\frac{y}{3}\right)+4\cdot \left(\frac{z}{4}\right)}{9}\ge {\left({\left(\frac{x}{2}\right)}^2{\left(\frac{y}{3}\right)}^3{\left(\frac{z}{4}\right)}^4\right)}^{\frac{1}{9}}$
${\left(\frac{1}{9}\right)}^9\ge \frac{x^2{y}^3{z}^4}{2^{10}{3}^3}\Rightarrow x^2{y}^3{z}^4\le \frac{2^{10}}{3^{15}}$