If $x,y,z \in \mathbb{R}$, $x^2+4y^2+16z^2=48, xy+4yz+2zx=24$, then find $x^2+y^2+z^2$. 
If $x,y,z \in \mathbb{R}$ are such that $x^2+4y^2+16z^2=48$ and $xy+4yz+2zx=24$, then find $x^2+y^2+z^2$.

I can find the answer if I find the value of $x+y+z$ and $xy+yz+zx$. But I don't know how to do that. I found that $$(x+2y+4z)^2=144 \implies x+2y+4z=±12$$
But I can't progress after this.

Now, by C-S $$144=3\cdot48=(1+1+1)(x^2+4y^2+16z^2)\geq(x+2y+4z)^2=144,$$
where the equality occurs for $$(1,1,1)||(x,2y,4z),$$
which gives  $$(x,y,z)=(4,2,1)$$ or
$$(x,y,z)=(-4,-2,-1)$$ and $$x^2+y^2+z^2=21.$$
