The answer is yes. The derivative in $\rho$ of the integral in question can be written as \begin{equation} J:=\int_{a}^{c}G(b)h(b)\,db, \end{equation} where \begin{equation} G(b):=(b-a)e^{\rho(b-a)}, \end{equation} \begin{equation} h(b):=F(b)^{n-3}\left[ (n-1)F(b)-(n-2)F(c)\right] f(b). \end{equation} It is enough to show that $J\ge0$.

Clearly, there is some $b_*\in[a,c]$ such that $h(b)\le0$ for $b\in[a,b_*)$ and $h(b)\ge0$ for $b\in[b_*,c]$. Moreover, the function $G$ is increasing, for each $\rho\ge0$. Therefore, \begin{equation} J:=\int_{[a,b_*)}G(b)h(b)\,db+\int_{[b_*,c])}G(b)h(b)\,db \end{equation} \begin{equation} \ge \int_{[a,b_*)}G(b_*)h(b)\,db+\int_{[b_*,c])}G(b_*)h(b)\,db \end{equation} \begin{equation} =G(b_*)\int_{[a,c]}h(u)\,du =G(b_*)F(a)^{n-2}[F(c)-F(a)]\ge0, \end{equation} as desired.

The answer is yes. The derivative in $\rho$ of the integral in question can be written as \begin{equation} J:=\int_{a}^{c}G(b)h(b)\,db, \end{equation} where \begin{equation} G(b):=(b-a)e^{\rho(b-a)}, \end{equation} \begin{equation} h(b):=F(b)^{n-3}\left[ (n-1)F(b)-(n-2)F(c)\right] f(b). \end{equation} It is enough to show that $J\ge0$.

Clearly, there is some $b_*\in[a,c]$ such that $h(b)\le0$ for $b\in[a,b_*)$ and $h(b)\ge0$ for $b\in[b_*,c]$. Moreover, the function $G$ is increasing, for each $\rho\ge0$. Therefore, \begin{equation} J=\int_{[a,b_*)}G(b)h(b)\,db+\int_{[b_*,c]}G(b)h(b)\,db \end{equation} \begin{equation} \ge \int_{[a,b_*)}G(b_*)h(b)\,db+\int_{[b_*,c]}G(b_*)h(b)\,db \end{equation} \begin{equation} =G(b_*)\int_{[a,c]}h(b)\,db =G(b_*)F(a)^{n-2}[F(c)-F(a)]\ge0, \end{equation} as desired.