show that inequality $(x-3)^4+(y-3)^4+(z-3)^4\ge 193$ Let $x,y,z\in R$,and such
$$xy+yz+xz=-1$$ show that
$$(x-3)^4+(y-3)^4+(z-3)^4\ge 193$$
it seem use Cauchy-Schwarz inequality to solve it?But I try sometime can't get this answer,even now I can't find this equality when $=$?

Let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative, and $xyz=w^3$.
Hence, the expression $\sum\limits_{cyc}(x-3)^4$ is a linear expression of $w^3$, 
which says that it gets a minimal value for an extremal value of $w^3$,
which happens for equality case of two variables.
Let $y=x$. Thus, $z=-\frac{1+x^2}{2x}$ and 
$$\min_{xy+xz+yz=-1}\sum_{cyc}(x-3)^4=\min\left(2(x-3)^4+\left(-\frac{1+x^2}{2x}-3\right)^4\right)=191.779...,$$
which occurs for $x_1=-0.12...$, where $x_1$ is a negative root of the following equation.
$$11x^6-90x^5+321x^4-180x^3-107x^2-18x-1=0.$$
