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?    
$108$ , $216$ , $405$ , $1048$

The problem can be restated as minimizing ${1 \over x^2} + {8 \over y^2} + {27 \over z^2}$ subject to $x + y + z = 1$. By Lagrange multipliers you seek an $(x,y,z)$ and $\lambda$ such that $-{2 \over x^3} = \lambda = -{16 \over y^3} = -{54 \over z^3}$. Dividing by $-2$ and taking cube roots leads to
$${1 \over x} = {2 \over y} = {3 \over z}$$
This can be rewritten as 
$$y = 2x,\,\,\,\,\,\,\,\,\,\,\, z = {3 \over 2} y = 3x$$
Plugging this back into the constraint gives
$$x + 2x + 3x = 1$$
So $x = {1 \over 6}$, whereupon $y = {1 \over 3}$ and $z = {1 \over 2}$. In this case, ${1 \over x^2} + {8 \over y^2} + {27 \over z^2} = 36 + 72 + 27*4 = 216$. This has to be the absolute minimum since the function ${1 \over x^2} + {8 \over y^2} + {27 \over z^2}$ goes to infinity as one approaches the hyperplanes $x = 0$, $y = 0$, or $z = 0$.
