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

Question

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.

Answer 1

$x^2+4y^2+16z^2=48 \cdots Eq.(1),\;\; xy+4yz+2zx=24 \cdots Eq.(2)$

From (1), $x^2+(2y)^2+(4z)^2=48 \cdots Eq.(3)$

$(2)\times4 \implies 4xy+16zy+8zx=96\cdots Eq.(4) $

$2\times Eq.(3)-Eq.(4)\implies 2[x^2+(2y)^2+(4z)^2]-2[2xy+8zy+4zx]=0 \implies (x-2y)^2+(2y-4z)^2+(x-4z)^2=0 \implies x=2y=4z.$

Now, $xy+4zy+2zx=24 \implies 3x^2=48\implies x^2=16 $.

Now, $x=4$, then $y=2,z=1$.

$\implies x^2+y^2+z^2=21$.

You will get the same answer even if you take $x=-4$.

Answer 2

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 $$x^2+y^2+z^2=21.$$

Answer 3

Let $a=x$, $b:=2y$, and $c:=4z$. Then, $$\begin{align}a^2+b^2+c^2&=x^2+4y^2+16z^2=48=2\cdot 24\\&=2\cdot(xy+4yz+2zx)=bc+ca+ab\,.\end{align}$$ It is well known that, for real numbers $a$, $b$, and $c$, $a^2+b^2+c^2=bc+ca+ab$ if and only if $a=b=c$. That is, $$x=2y=4z\,.$$ Since $x^2+4y^2+16z^2=48$, we get $$(x,y,z)=\pm(4,2,1)\,.$$ That is, $$x^2+y^2+z^2=16+4+1=21\,.$$

Answer 4

$$0 = x^2 + 4 y^2 + 16 z^2 - 2 (4yz + 2zx + xy) = (x-y-2z)^2 + 3 (y-2z)^2 $$ So, as noted $y = 2z$ and $x = y+2z = 4z,$ so $$ 48 = x^2 + 4 y^2 + 16 z^2 = 16 z^2 + 16 z^2 + 16 z^2 = 48 z^2 $$ and $z = \pm 1.$