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.