Ratio of geodesic segments on the sphere Let $\mathbb{S}^2$ be the unit sphere. Let $0<\lambda<1$ be fixed. What is the smallest number $0<\mu<1$ (depending on $\lambda$) such that for any three points $A,B,C\in \mathbb{S}^2$, with $d(C,A)\le \pi/2, d(C,B)\le \pi/2$, one has $d(A_{\lambda},B_{\lambda})\le \mu d(A,B)$. Here $A_{\lambda}$ and $B_{\lambda}$ are points on the geodesic paths $[A,C]$ and $[B,C]$ such that $d(A_{\lambda},C)=\lambda d(A,C)$ and $d(B_{\lambda},C)=\lambda d(B,C)$. ($d$ is the geodesic distance.)
My conjecture is $\sin(\lambda\pi/2)$. Is this correct?
 A: Yes, this is correct.
Since you've found this result, I'll assume that you figured out the case where $A$ and $B$ are at equal distance from $C$, and it remains only to be shown that this is the worst case. Let me know if you'd like me to write more on the simpler case.
We can write $A$ and $B$ in spherical coordinates with $C$ at one pole and $A$ at $\phi=0$:
$$A=\pmatrix{\sin\theta_A\\0\\\cos\theta_A},\quad B=\pmatrix{\sin\theta_B\cos\phi\\\sin\theta_B\sin\phi\\\cos\theta_B}\;.$$
Then
$$A_\lambda=\pmatrix{\sin\lambda\theta_A\\0\\\cos\lambda\theta_A},\quad B_\lambda=\pmatrix{\sin\lambda\theta_B\cos\phi\\\sin\lambda\theta_B\sin\phi\\\cos\lambda\theta_B}\;,$$
and the geodesic distances are the arccosines of the scalar products:
$$
\begin{eqnarray}
d(A,B)&=&\arccos\left(\sin\theta_A\sin\theta_B\cos\phi+\cos\theta_A\cos\theta_B\right)\;,\\
d(A_\lambda,B_\lambda)&=&\arccos\left(\sin\lambda\theta_A\sin\lambda\theta_B\cos\phi+\cos\lambda\theta_A\cos\lambda\theta_B\right)\;.
\end{eqnarray}
$$
We can rewrite them in terms of $\Sigma=\theta_A+\theta_B$ and $\Delta=\theta_A-\theta_B$:
$$
\begin{eqnarray}
d(A,B)&=&\arccos\frac12\left((1-\cos\phi)\cos\Sigma+(1+\cos\phi)\cos\Delta
\right)\;,\\
d(A_\lambda,B_\lambda)&=&
\arccos\frac12\left((1-\cos\phi)\cos\lambda\Sigma+(1+\cos\phi)\cos\lambda\Delta\right)
\;.
\end{eqnarray}
$$
To see where the desired inequality is most strongly violated, we can differentiate the discrepancy with respect to $\Delta$:
$$\frac{\partial}{\partial\Delta}\left(d(A_\lambda,B_\lambda)-\mu d(A,B)\right)=\frac12(1+\cos\phi)\left(\frac{\lambda\sin\lambda\Delta}{\sin d(A_\lambda,B_\lambda)}-\mu\frac{\sin\Delta}{\sin d(A,B)}\right)\;.$$
Since we already know $\mu\ge\sin(\lambda\pi/2)\ge\lambda$, we must have $d(A_\lambda,B_\lambda)\gt\lambda d(A,B)$ for the inequality to be violated, and thus $\sin d(A_\lambda,B_\lambda)\gt\lambda \sin d(A,B)$. As all the quantities are non-negative (since $\theta\le\pi/2$ and $0\lt\lambda\lt1$), we can bound the derivative thus:
$$
\begin{eqnarray}
\frac{\partial}{\partial\Delta}\left(d(A_\lambda,B_\lambda)-\mu d(A,B)\right)
&\le&
\frac12(1+\cos\phi)\left(\frac{\lambda\sin\lambda\Delta}{\lambda\sin d(A,B)}-\mu\frac{\sin\Delta}{\sin d(A,B)}\right)\\
&=&
\frac12\frac{1+\cos\phi}{\sin d(A,B)}\left(\sin\lambda\Delta-\mu\sin\Delta\right)\\
&\le&
\frac12\frac{1+\cos\phi}{\sin d(A,B)}\left(\sin\lambda\Delta-\sin\frac{\lambda\pi}2\sin\Delta\right)\\
&\le&0\;.
\end{eqnarray}
$$
Thus, the discrepancy is non-increasing wherever it is non-negative, and hence for it to be positive anywhere, it would have to be positive at $\Delta=0$. It therefore suffices to treat that case, which leads to the result you found.
