Hadamard squared-distance inequality ($d(z,\gamma (t){)}^2\le (1-t)d(z,x{)}^2+td(z,y{)}^2-t(1-t)d(x,y{)}^2$)

For the constant-speed geodesic $\gamma$ from $x$ to $y$ in a Hadamard space, the displayed inequality follows by repeatedly applying the midpoint inequality and passing from dyadic parameters to all $t\in [0,1]$. Applying it on two sides of a triangle gives $d(x_s,y_t{)}^2\le s^2{A}^2+t^2{B}^2-st(A^2+B^2-C^2)$, where $A=d(a,x)$, $B=d(a,y)$ and $C=d(x,y)$. The right side is the squared comparison distance by the Euclidean cosine rule.