Slim triangles imply the gromov product inequality

Put $\alpha =\min \{(x\mathrm{\mid }y{)}_o,(y\mathrm{\mid }z{)}_o\}$. For any point $v\in [x,y]$, the inequalities $d(o,x)\le d(o,v)+d(v,x)$ and $d(o,y)\le d(o,v)+d(v,y)$ add to give $d(o,v)\ge (x\mathrm{\mid }y{)}_o$. The same argument applies to $[y,z]$.

If $\alpha \le \delta$, nonnegativity gives the claimed lower bound. Otherwise fix $0\le t<\alpha -\delta$ and $0<h<\alpha -\delta -t$. Choose points $x_t,y_t,z_t$ at distance $t$ from $o$ on three chosen radial sides. These points exist since each radial length is at least $\alpha$. Slimness and the infimum convention supply a point within distance less than $\delta +h$ of $x_t$ on $[o,y]\cup [x,y]$. It cannot lie on $[x,y]$, whose points have distance at least $\alpha >t+\delta +h$ from $o$. Let it be $v\in [o,y]$. Then $\mathrm{\mid }d(o,v)-t\mathrm{\mid }<\delta +h$, so $d(x_t,y_t)<2(\delta +h)$. Applying the same reasoning to $y_t$ in triangle $(o,y,z)$ gives $d(y_t,z_t)<2(\delta +h)$.

Thus $d(x_t,z_t)<4(\delta +h)$ and the joining route through them gives $d(x,z)\le d(o,x)-t+4(\delta +h)+d(o,z)-t$. Expanding the product yields $(x\mathrm{\mid }z{)}_o\ge t-2(\delta +h)$. This is true for every sufficiently small positive $h$, so $(x\mathrm{\mid }z{)}_o\ge t-2\delta$. Otherwise a sufficiently small $h$ would contradict the strict gap.

Letting $t$ approach $\alpha -\delta$ from below in the same elementary real inequality gives $(x\mathrm{\mid }z{)}_o\ge \alpha -3\delta$. When $\delta =0$ and $\alpha >0$, the same positive $h$ and $t<\alpha$ argument applies; when both vanish step 2.1 applies. All side selections are finite and all near-point witnesses are used one at a time, so no AC or closest-point attainment was used.