$\definecolor{A}{RGB}{0,90,255}\color{A}\fbox{Solution}$ Multiply through $k>0$ every variable. Take derivative on $k$ and conclude inequality is true for all $k>0$ (in particular for $k=1$) iff it holds for $k=\frac{3}{x+y+z}$, that is iff $$(x+y+z)^3\ge 54xyz.$$Substitute $a=\frac{2x}{x+y+z},...$ Then $a+b+c=2,\ ab+bc+ca=1$ and inequality becomes $$\frac4{27}\ge abc.$$Transform equations into $b+c=2-a,\ bc=1-a(b+c)=(1-a)^2$ and now everything is clear by AM-GM. #1770

parmenides51 wrote: Let $x, y, z$ be positive real numbers satisfy the condition $x^2 +y^2 + z^2 = 2(x y + yz + z x)$. Prove that $x + y + z + \frac{1}{2x yz} \ge 4$ ON SOME CYCLIC HOMOGENEOUS POLYNOMIAL INEQUALITIES OF DEGREE FOUR IN REAL VARIABLES UNDER CONSTRAINTS