**EDIT:** I misread the sense of your inequality at my first read. It seems to me that the inequality you are asking for cannot be true, not even locally. In the spirit of comparison geometry, upper (resp. lower) bounds on curvature imply lower (resp. upper) bounds on the distance in terms of comparison triangles.

1) A similar result, but with reversed inequality holds and follows immediately from Berger's comparison theorem, **and provided that the triangle $PQR$ is sufficiently small** (I think that it is enough to be contained in a ball with center $P$ of radius equal to the injectivity radius from $P$).

The result I'm referring to is a basic comparison theorem, number 1.29 in:

*Cheeger, Jeff; Ebin, David G.*, Comparison theorems in Riemannian geometry, North-Holland Mathematical Library. Vol. 9. Amsterdam-Oxford: North-Holland Publishing Company; New York: American Elsevier Publishing Company, Inc. VIII, 174 p. Dfl. 50.00; $ 19.25 (1975). ZBL0309.53035.

More precisely, if $PQR$ is sufficiently small and $\mathrm{Sec} \leq k$, then

$$ d(Q_t,R_t) \geq \tilde{d}_k (\tilde{Q}_t,\tilde{R}_t) $$

On the right hand side of your inequality, $\tilde{d}_k$ is the distance function between $\tilde{Q}_t$ and $\tilde{R}_t$ in the model space with constant curvature equal to $k$, where $\tilde{Q},\tilde{R}, \tilde{P}$ is a comparison triangle on the model space with sides equal to the geodesic triangle with $Q,R,P$ as vertex on the originale space.

In the very special case $k=0$ is as you say, and $\tilde{d}_k(\tilde{Q}_t,\tilde{R}_t) = t \tilde{d}_k(\tilde{Q},\tilde{R}) = t d(Q,R)$.

2) If you assume, instead, a lower curvature bound you have Toponogov's theorem, which is an inequality in the sense you ask. See e.g. Theorem 2.2, statement (B) in the aforementioned reference (in this case, you don't need your triangle PQR to be small).

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