[proofplan] The proof is a direct conversion of Hamilton's trace Harnack inequality into an inequality along a prescribed curve. We compute the derivative of $\tau R(\gamma(\tau),\tau)$ by the product rule and the chain rule on $M \times (0,T]$. Then we apply the assumed trace Harnack inequality at the point $(\gamma(\tau),\tau)$ with the special choice $V=\dot{\gamma}(\tau)/2$, which makes the spatial gradient term match the curve derivative term. [/proofplan]

[step:Compute the derivative of scalar curvature along the curve]Define the scalar function $\rho:(s,t) \to \mathbb{R}$ by $\rho(\tau)=R(\gamma(\tau),\tau)$. Since $R$ is smooth on $M \times (0,T]$ and $\gamma$ is smooth, $\rho$ is smooth on $(s,t)$. By the chain rule for the map $\tau \mapsto (\gamma(\tau),\tau)$, \begin{align*} \rho'(\tau) = g(\tau)_{\gamma(\tau)} \big(\nabla^{g(\tau)} R(\gamma(\tau),\tau),\dot{\gamma}(\tau)\big) + \partial_\tau R(\gamma(\tau),\tau). \end{align*} Therefore the product rule gives \begin{align*} \frac{d}{d\tau}\big(\tau R(\gamma(\tau),\tau)\big)=R(\gamma(\tau),\tau)+\tau \rho'(\tau). \end{align*} Substituting the formula for $\rho'(\tau)$ gives \begin{align*} \frac{d}{d\tau}\big(\tau R(\gamma(\tau),\tau)\big)=R(\gamma(\tau),\tau)+\tau\partial_\tau R(\gamma(\tau),\tau)+\tau g(\tau)_{\gamma(\tau)}\big(\nabla^{g(\tau)} R(\gamma(\tau),\tau),\dot{\gamma}(\tau)\big). \end{align*}[/step]

[guided]We first isolate the ordinary one-variable function obtained by evaluating scalar curvature along the curve. Define the scalar function $\rho:(s,t) \to \mathbb{R}$ by $\rho(\tau)=R(\gamma(\tau),\tau)$. This function depends on $\tau$ in two ways: the point $\gamma(\tau)$ moves in $M$, and the metric time variable in $R(\cdot,\tau)$ also changes. Since $R$ is smooth on $M \times (0,T]$ and $\gamma$ is smooth, the chain rule applies to the smooth curve $\Gamma:(s,t) \to M \times (0,T]$ defined by $\Gamma(\tau)=(\gamma(\tau),\tau)$. The spatial part of the derivative is the directional derivative of $R(\cdot,\tau)$ in the direction $\dot{\gamma}(\tau)$, and by the definition of the gradient with respect to $g(\tau)$ this directional derivative is \begin{align*} g(\tau)_{\gamma(\tau)} \big(\nabla^{g(\tau)} R(\gamma(\tau),\tau),\dot{\gamma}(\tau)\big). \end{align*} The time part is $\partial_\tau R(\gamma(\tau),\tau)$. Hence \begin{align*} \rho'(\tau) = g(\tau)_{\gamma(\tau)} \big(\nabla^{g(\tau)} R(\gamma(\tau),\tau),\dot{\gamma}(\tau)\big) + \partial_\tau R(\gamma(\tau),\tau). \end{align*} Now apply the ordinary product rule to $\tau \rho(\tau)$: \begin{align*} \frac{d}{d\tau}\big(\tau R(\gamma(\tau),\tau)\big)=\frac{d}{d\tau}\big(\tau \rho(\tau)\big). \end{align*} The product rule gives \begin{align*} \frac{d}{d\tau}\big(\tau \rho(\tau)\big)=\rho(\tau)+\tau \rho'(\tau). \end{align*} Substituting the definitions of $\rho(\tau)$ and $\rho'(\tau)$ yields \begin{align*} \frac{d}{d\tau}\big(\tau R(\gamma(\tau),\tau)\big)=R(\gamma(\tau),\tau)+\tau\partial_\tau R(\gamma(\tau),\tau)+\tau g(\tau)_{\gamma(\tau)}\big(\nabla^{g(\tau)} R(\gamma(\tau),\tau),\dot{\gamma}(\tau)\big). \end{align*} This is the exact expression that Hamilton's trace Harnack inequality will estimate from below.[/guided]

[step:Apply the trace Harnack inequality with half the curve velocity] Fix $\tau \in (s,t)$, and define the tangent vector \begin{align*} V_\tau:=\frac{1}{2}\dot{\gamma}(\tau)\in T_{\gamma(\tau)}M. \end{align*} The assumed trace Harnack inequality applies at the point $x=\gamma(\tau)$ and time $\tau$ with this vector $V_\tau$. Thus \begin{align*} 0\le \partial_\tau R(\gamma(\tau),\tau)+2\,g(\tau)_{\gamma(\tau)}\big(\nabla^{g(\tau)} R(\gamma(\tau),\tau),V_\tau\big)+2\,\operatorname{Ric}_{g(\tau),\gamma(\tau)}(V_\tau,V_\tau)+\frac{1}{\tau}R(\gamma(\tau),\tau). \end{align*} Using $V_\tau=\frac{1}{2}\dot{\gamma}(\tau)$, bilinearity of the metric, and bilinearity of the Ricci tensor, this becomes \begin{align*} 0\le \partial_\tau R(\gamma(\tau),\tau)+g(\tau)_{\gamma(\tau)}\big(\nabla^{g(\tau)} R(\gamma(\tau),\tau),\dot{\gamma}(\tau)\big)+\frac{1}{2}\operatorname{Ric}_{g(\tau),\gamma(\tau)}\big(\dot{\gamma}(\tau),\dot{\gamma}(\tau)\big)+\frac{1}{\tau}R(\gamma(\tau),\tau). \end{align*} Multiplying by $\tau>0$ gives \begin{align*} R(\gamma(\tau),\tau)+\tau\partial_\tau R(\gamma(\tau),\tau)+\tau g(\tau)_{\gamma(\tau)}\big(\nabla^{g(\tau)} R(\gamma(\tau),\tau),\dot{\gamma}(\tau)\big)\ge -\frac{\tau}{2}\operatorname{Ric}_{g(\tau),\gamma(\tau)}\big(\dot{\gamma}(\tau),\dot{\gamma}(\tau)\big). \end{align*} [/step]

[step:Identify the left hand side with the curve derivative] By the derivative formula obtained above, the left hand side of the last inequality is exactly \begin{align*} \frac{d}{d\tau}\big(\tau R(\gamma(\tau),\tau)\big). \end{align*} Therefore, for every $\tau \in (s,t)$, \begin{align*} \frac{d}{d\tau}\big(\tau R(\gamma(\tau),\tau)\big)\ge -\frac{\tau}{2}\operatorname{Ric}_{g(\tau),\gamma(\tau)}\big(\dot{\gamma}(\tau),\dot{\gamma}(\tau)\big). \end{align*} At $\tau=s$, the same chain rule and product rule apply to the right-hand derivative of the restriction of $\rho$ to $[s,t]$; at $\tau=t$, they apply to the left-hand derivative. The trace Harnack inequality also applies at both endpoint times because $0<s<t\le T$, so $s,t\in(0,T]$. Hence the same inequality holds with the corresponding one-sided derivatives at the endpoints. This proves the claimed scalar curvature Harnack inequality along $\gamma$. [/step]