Mathematics - Sequence & Series Question with Solution | TestHub

A.M. $\ge$ G.M. on $\frac{x}{2},\frac{x}{2},\frac{y}{3},\frac{y}{3},\frac{y}{3},\frac{z}{2},\frac{z}{2}:\frac{2}{7}\ge {\left(\frac{x^2{y}^3{z}^2}{2^2\cdot 3^3\cdot 2^2}\right)}^{1/7}$, so $x^2{y}^3{z}^2\le \frac{2^{11}\cdot 3^3}{7^7}$ (about 0.067, so the maximum is not 1). $\frac{x+y+z}{3}\ge (xyz{)}^{1/3}\Rightarrow xyz\le \frac{8}{27}$. By Cauchy-Schwarz, $(x+y+z)\left(\frac{4}{x}+\frac{1}{y}+\frac{9}{z}\right)\ge (2+1+3{)}^2=36$, so $\frac{4}{x}+\frac{1}{y}+\frac{9}{z}\ge 18$, with equality attainable.