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

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 $p=1,q=5$)
