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?    
$108$ , $216$ , $405$ , $1048$
 A: The function $t\mapsto t^3$ is convex on ${\mathbb R}_{>0}$. 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 $(6,3,2)$.
A: As $x,y,z$ are +ve real, we can set $\frac 1x=\sin^2A,\frac1y+\frac1z=\cos^2A$
again, $\frac1{y\cos^2A}+\frac1{z\cos^2A}=1$
we  can put $\frac1{y\cos^2A}=\cos^2B, \frac1{z\cos^2A}=\sin^2B\implies y=\cos^{-2}A\cos^{-2}B,z=\cos^{-2}A\sin^{-2}B$
So, $$x^2+8y^2+27z^2=\sin^{-4}A+\cos^{-4}A(8\cos^{-4}B+27\sin^{-4}B)$$
We need $8\cos^{-4}B+27\sin^{-4}B$ to be minimum for the minimum value of $x^2+8y^2+27z^2$
Let $F(B)=p^3\cos^{-4}B+q^3\sin^{-4}B$ where $p,q$ are positive real numbers.
then $F'(B)=p^3(-4)\cos^{-5}B(-\sin B)+q^3(-4)\sin^{-5}B(\cos B)$
for the extreme values of $F(B),F'(B)=0\implies (p\sin^2B)^3=(q\cos^2B)^3\implies \frac{\sin^2B}q=\frac{\cos^2B}p=\frac1{p+q}$
Observe that $F(B)$ can not have any finite maximum value.
So, $F(B)_{min}=\frac{p^3}{\left(\frac p{p+q}\right)^2}+\frac{q^3}{\left(\frac q{p+q}\right)^2}=(p+q)^3$
So, the minimum value of $8\cos^{-4}B+27\sin^{-4}B$ is $(2+3)^3=125$ (Putting $p=2,q=3$)
So, $$x^2+8y^2+27z^2=\sin^{-4}A+\cos^{-4}A(8\cos^{-4}B+27\sin^{-4}B)\ge \sin^{-4}A+125\cos^{-4}A\ge (1+5)^3=216$$   (Putting $p=1,q=5$)
A: $$(x^2+8y^2+27z^2)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 \ge (1+2+3)^3 = 216$$
by Holder's inequality. It's clear that the equality can hold.
A: Let $F=1/x+1/y+1/z-1$ and $G=x^2+8y^2+27z^2$. You seek min of $G$ with constraint $F=0.$  Scalar multiples of the gradients are 
$$-\nabla F=(1/x^2,1/y^2,1/z^2),$$
$$(1/2)\nabla G=(x,8y,27z).$$
Using Lagrange multipliers this means at a critical point the following are zero:
$$F_xG_y-F_yG_x=-(x-2y)(x^2+2xy+4y^2),$$
$$F_yG_z-G_yF_z=-(2y-3z)(4y^2+6yz+9z^2),$$
$$F_xG_z-G_xF_z=-(x-3z)(x^2+3xz+9z^2).$$
Since the variables are positive this means at a critical point $x=2y,\ 2y=3z,\ x=3z$ which with the constraint $F=0$ gives the single critical point $(6,3,2)$. 
At $(6,3,2)$ we have $G=216.$ This is one of the choices in the question. Note that at $(2,3,6)$ the value is $G=1048,$ one of the other choices, leading one to conjecture the test makers switched the coordinates to come up with another multiple choice possibility.
NOTE: I haven't rigorously proved that $G$ takes on its absolute minimum at the critical point $(6,3,2)$. But it seems likely, in that moving one of the variables near $0$ causes one of the others to get large, making $G$ become large since the variables are squared in $G$. If I can rigorize I'll add more to this answer. Or if someone else can do so that would be good.
A: 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 $z = 0$.
A: 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.
