Almost 60 followers problem

Happy Victory Day, Comrade!

$(2x-2y)^2+(4y-z)^2+(4x-z)^2=20x^2+20y^2+2z^2-8xy-8yz-8xz\geq 0$ so that $10x^2+10y^2+z^2\geq 4xy+4yz+4xz=4$ . The minimum is $\boxed{4}$ , attained when $x=y=\frac{1}{3}$ and $z=\frac{4}{3}$

AM-GM? I used Cauchy-schwarz TT.TT

$(1^2+1^2+2^2)((2x)^2+(2y)^2+z^2) \geq (2x+2y+2z)^2$

$(6)(4x^2+4y^2+z^2) \geq 4(x+y+z)^2$

$4x^2+4y^2+z^2 \geq \frac{2}{3} (x^2+y^2+z^2+2(xy+yz+xz))$

$4x^2+4y^2+z^2 \geq \frac{2}{3} (x^2+y^2+z^2+2)$ (Since $xy+yz+xz=1$ )

$\frac{10x^2+10y^2+z^2}{3} \geq \frac{4}{3}$ (Transposition)

$10x^2+10y^2+z^2 \geq \boxed {4}$

I first tried symmetry in $x$ , $y$ , and $z$ , but it didn't work -- presumably because the two equations are not mutually symmetric in $z$ (substituting just $y=x$ gives the expression as $20x^2+z^2$ and the equation as $x^2+2xz=1$ ). But if you want to go this way, you could take just the symmetry in $x$ and $y$ and proceed from there:

Solve the previous equation for $z$ to get $z=\dfrac{1-x^2}{2x}$ and $z^2=\dfrac{1-2x^2+x^4}{4x^2}=\frac{1}{4}x^{-2}-\frac{1}{2}+\frac{1}{4}x^2$ .

Substitute into the expression to get $20x^2+\frac{1}{4}x^{-2}-\frac{1}{2}+\frac{1}{4}x^2=\frac{1}{4}x^{-2}-\frac{1}{2}+\frac{81}{4}x^2$ .

Now, by AM-GM, $\frac{1}{4}x^{-2}-\frac{1}{2}+\frac{81}{4}x^2\geq-\frac{1}{2}+2\sqrt{\frac{1}{4}x^{-2}*\frac{81}{4}x^2}=-\frac{1}{2}+2(\frac{9}{4})=\boxed{4}$ .

Equality occurs when $\frac{81}{4}x^2=\frac{1}{4}x^{-2}$ , i.e. $x^4=\frac{1}{81}$ , i.e. $x=\frac{1}{3}$ , $y=\frac{1}{3}$ , $z=\frac{4}{3}$ .

Let $x=\kappa , y=\kappa , z= \sqrt{10}\kappa$ .

The above ratio is taken so as to have symmetry to get minimum.

Thus $\kappa= 1/(1+2\sqrt{10})$ .

Now putting in given expression min value is

$\approx 4.095$ .