For any real numbers $a$, the numbers $a-1,\,a^2-1,\,a^3-1,\,a^5-1$ are of the same sign.
Therefore
$(a-1)(a^2-1)\ge 0$, $(a^2-1)(a^3-1)\ge 0$ and $(a^2-1)(a^5-1)\ge 0$.
i.e.
$x^3-x^2-x+1\ge 0$
$y^5-y^3-y^2+1\ge 0$
$z^7-z^5-z^2+1\ge 0$
So it follows that
$x^3-x+5\ge x^2+4$, $y^5-y^3+5\ge y^2+4$ and $z^7-z^5+5\ge z^2+4$
Multiplying these inequalities gives
$(x^3-x+5)(y^5-y^3+5)(z^7-z^5+5)\ge(x^2+4)(y^2+4)(z^2+4) \tag{1}$
We will prove that
$(x^2+4)(y^2+4)(z^2+4)\ge 25(xy+yz+xz+2) \tag{2}$
We have
$\begin{align*}(x^2+4)(y^2+4)(z^2+4)&=x^2y^2z^2+4(x^2y^2+y^2z^2+z^2x^2)+16(x^2+y^2+z^2)+64\\&=x^2y^2z^2+(x^2+y^2+z^2)+2+4(x^2y^2+y^2z^2+z^2x^2+3)+15(x^2+y^2+z^2)+50---(3)\end{align*}$
By the inequalities
$(x-y)^2+(y-z)^2+(z-x)^2\ge 0$ and
$(xy-1)^2+(yz-1)^2+(zx-1)^2\ge 0$ we obtain
$x^2+y^2+z^2\ge xy+yz+zx\tag{4}$ and
$x^2y^2+y^2z^2+z^2x^2+3\ge 2(xy+yz+zx)\tag{5}$
We will prove that
$x^2y^2z^2+(x^2+y^2+z^2)+2\ge 2(xy+yz+zx)\tag{6}$
Note that if we have $a,\,b,\,c>0$ then $3abc+a^3+b^3+c^3\ge 2((ab)^{\dfrac{3}{2}}+(bc)^{\dfrac{3}{2}}+(ca)^{\dfrac{3}{2}})$.
Its proof followed by Schur's inequality and AM-GM inequality.
For $a=x^{\dfrac{2}{3}}$, $b=y^{\dfrac{2}{3}}$ and $c=z^{\dfrac{2}{3}}$, we deduce
$3(xyz)^{\dfrac{2}{3}}+x^2+y^2+z^2\ge 2(xy+yz+zx)$
Therefore, it suffices to prove that
$x^2y^2z^2+2\ge 3(xyz)^{\dfrac{2}{3}}$, which follows immediately by $AM>GM$.
Thus we have proved inequality (6).
Now, by (3), (4), (5) and (6) we obtain inequality (2).
Finally by (1) and (2) and since $xy+yz+zx=3$ we obtain the required inequality.
Equality occurs iff $x=y=z=1$.
