Equality holds when $x\to \frac{\sqrt{3}+5}{1-2 \left(1-\sqrt{3}\right)},y\to -2,z\to 1-\sqrt{3}.$

SBM wrote: Equality holds when $x\to \frac{\sqrt{3}+5}{1-2 \left(1-\sqrt{3}\right)},y\to -2,z\to 1-\sqrt{3}.$ To simplify, $x=1+\sqrt{3}$.

sqing wrote: For any real numbers $x, y, z$ with $xyz + x + y + z = 4, $show that $$(yz + 6)^2 + (zx + 6)^2 + (xy + 6)^2 \geq 8 (xyz + 5).$$ Let $xy+yz+zx=q$ and $xyz=r.$ So $x+y+z=4-r.$ So it reduces to $$(q+6)^2+2(r-4)^2\geq0,$$which is true. Equality for the roots of the cubic $u^3-6u-4=0,$ i.e. the permutations of $\left(-2,1-\sqrt3,1+\sqrt3\right).$

sqing wrote: For any real numbers $x, y, z$ with $xyz + x + y + z = 4, $show that $$(yz + 6)^2 + (zx + 6)^2 + (xy + 6)^2 \geq 8 (xyz + 5).$$

Attachments:

$$(yz + 6)^2 + (zx + 6)^2 + (xy + 6)^2 \geq 8 (xyz + 5).$$$$xyz=A$$$$xy+yz+zx=B$$$$x+y+z=4-A$$$$B^2-2AC+12B+68-8A\ge0$$$$(B+6)^2+2(A-4)^2\ge0$$

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