The gromov product inequality implies the four point condition

For an ordered quadruple $(o,x,y,z)$ write $A=d(o,y)+d(x,z)$, $B=d(o,z)+d(x,y)$ and $C=d(o,x)+d(y,z)$, and put $R=d(o,x)+d(o,y)+d(o,z)$. Then $2(x\mathrm{\mid }z{)}_o=R-A$, $2(x\mathrm{\mid }y{)}_o=R-B$, and $2(y\mathrm{\mid }z{)}_o=R-C$. Consequently the product inequality for this ordered quadruple is exactly $A\le \max \{B,C\}+2\kappa$, since $\min \{R-B,R-C\}=R-\max \{B,C\}$.

Suppose the product condition holds for all ordered quadruples. Permuting $x,y,z$ in step 1.1 gives the three inequalities bounding each of $A,B,C$ by the maximum of the other two plus $2\kappa$. Apply the inequality with the largest sum on its left: the maximum on its right is the second-largest, including ties. This proves the four-point condition.

Conversely suppose the four-point condition holds. For every ordered quadruple, if $A$ is largest it is at most the second-largest plus $2\kappa$, while if it is not largest it is already at most $\max \{B,C\}$. Thus $A\le \max \{B,C\}+2\kappa$ in either case. Step 1.1 recovers the product inequality at the arbitrary basepoint $o$. The computations remain valid when points coincide or $\kappa =0$.