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 function $t\mapsto t^3$ is convex on ${\mathbb R}_{&gt;0}$. Given that $${1\over x}+{1\over y}+{1\over z}=1\qquad(*)$$ we therefore have by Jensen's inequality
$$x^2+8y^2+27 z^2={1\over x} x^3 + {1\over y}(2y)^3+{1\over z}(3z)^3 \geq\left({1\over x} x+ {1\over y}2y +{1\over z} 3z\right)^3 =216\ ,$$
where equality holds iff in addition to $(*)$ we have $x=2y=3z$. It follows that the minimum value we are looking for is $216$ and that it is taken at the point $(6,3,2)$.
x:y:z=3:2:1 then x=6,y=6/3 and z=6/2 ('.' 1^3=1, 2^3=8, 3^3=27 and 1+2+3=6)
substitute in the equation
6^2+8(3^2)+27(2^2)=216
u can exchange the value of x y z but he has asked for minimum value for expression.so taking these value exactly for x y and z would be appropriate.