If x,y,z are positive reals, then the minimum value of $x^2+8y^2+27z^2$ where $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ is what If $x,y, z$ are positive reals, then the minimum value of $x^2+8y^2+27z^2$ where $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ is    what?    
$108$ , $216$ , $405$ , $1048$

$$(x^2+8y^2+27z^2)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 \ge (1+2+3)^3 = 216$$
by Holder's inequality. It's clear that the equality can hold.
