= Hadamard squared-distance inequality
{c}
{title2=$d(z,\gamma(t))^2\leq(1-t)d(z,x)^2+td(z,y)^2-t(1-t)d(x,y)^2$}
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 $d(x_s,y_t)^2\leq s^2A^2+t^2B^2-st(A^2+B^2-C^2)$, 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.
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