= Solution
The comparison triangle must have third side $d(x,y)$: the printed $d(y,z)$ contains an undefined $z$. Use $A=d(a,x)$, $B=d(a,y)$ and $C=d(x,y)$.
First extend the midpoint inequality along any constant-speed <geodesic> $\gamma$ from $p$ to $q$. If $L=d(p,q)$, the function $d(z,\gamma(t))^2-L^2t^2$ is midpoint-convex by (a). Iteration proves its convexity bound for dyadic $t$, and <continuity> then gives the <Hadamard squared-distance inequality>
$$
d(z,\gamma(t))^2\leq(1-t)d(z,p)^2+td(z,q)^2-t(1-t)L^2.
$$
Apply this first to $[a,x]$ with $z=y_t$, and then to $[a,y]$ with $z=x$:
$$
d(x_s,y_t)^2\leq(1-s)t^2B^2+s\bigl[(1-t)A^2+tC^2-t(1-t)B^2\bigr]
-s(1-s)A^2.
$$
Simplification gives
$$
\boxed{d(x_s,y_t)^2\leq s^2A^2+t^2B^2-st(A^2+B^2-C^2).}
$$
In a Euclidean comparison triangle, the cosine rule gives the scalar product of the two <vectors> from the common vertex as $(A^2+B^2-C^2)/2$. The squared distance between their multiples $s$ and $t$ is exactly the right side displayed above. Taking nonnegative square roots proves the required triangle comparison, including degenerate Euclidean triangles.
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