Prove a certain particle is always ahead of the other particle. Suppose that the motion of classical particles $a$ and $b$ are governed by the following equations
$$
    \ddot{x}_a + f \dot{x}_a = F(x_a)
$$
and 
$$
    \ddot{x}_b + f \dot{x}_b = G(x_b).
$$
where $f >0$. 
The initial condition of both particles are set to be the same as $x_{a,b} = 0$ and $\dot{x}_{a,b} = 0$. 
Also we assume that $F(x) > 0$,  $G(x) > 0$ and $F(x) > G(x)$.
Therefore, the force to the particle $a$ is always larger than that to the particle $b$. 
Under those conditions, I have a feeling that $x_a(t) > x_b(t)$, $\dot{x}_a(t) > \dot{x}_b(t)$ or $\dot{x}_a(x_a) > \dot{x}_b(x_b)$ should hold. 
However, I cannot prove any of those myself. 
Please give me some hints or a complete proof whatever might be helpful for me. 
thank you.
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
We'll consider
$\ds{\ddot{\rm x}\pars{t} + f\dot{\rm x}\pars{t}={\cal F}\pars{x\pars{t}}}$ which leads to:

$$
\totald{\bracks{\expo{ft}\dot{\rm x}\pars{t}}}{t}=\expo{ft}{\cal F}\pars{x\pars{t}}
\ \imp\ \expo{ft}\dot{\rm x}\pars{t}
=\int_{0}^{t}\expo{fs}{\cal F}\pars{x\pars{s}}\,\dd s
$$

$$
\dot{\rm x}\pars{t}
=\expo{-ft}\int_{0}^{t}\expo{fs}{\cal F}\pars{x\pars{s}}\,\dd s\ \imp\
{\rm x}\pars{t}
=\int_{0}^{t}\expo{-fs'}\int_{0}^{s'}\expo{fs}{\cal F}\pars{x\pars{s}}
\,\dd s\,\dd s'
$$

$$
{\rm x}\pars{t}
=\int_{0}^{t}\expo{fs}{\cal F}\pars{x\pars{s}}\int_{s}^{t}\expo{-fs'}
\,\dd s'\,\dd s
=\int_{0}^{t}\expo{fs}{\cal F}\pars{x\pars{s}}
\pars{\expo{-ft} - \expo{-fs} \over -f}\,\dd s
$$

\begin{align}
\color{#c00000}{{\rm x}\pars{t}}&={1 \over f}\int_{0}^{t}\bracks{%
1 - \expo{-f\pars{t - s}}}{\cal F}\pars{x\pars{s}}\,\dd s
\\[5mm]
\color{#c00000}{\dot{\rm x}\pars{t}}&=\int_{0}^{t}
\expo{-f\pars{t - s}}{\cal F}\pars{x\pars{s}}\,\dd s
\end{align}

Then
  \begin{align}
\color{#c00000}{{\rm x_{a}}\pars{t} - {\rm x_{b}}\pars{t}}&
={1 \over f}\int_{0}^{t}\bracks{1 - \expo{-f\pars{t - s}}}
\bracks{{\rm F}\pars{{\rm x_{a}}\pars{s}} - {\rm G}\pars{{\rm x_{b}}\pars{s}}}
\,\dd s
\\[5mm]
\color{#c00000}{\dot{\rm x}_{\rm a}\pars{t} - \dot{\rm x}_{\rm b}\pars{t}}&
=\int_{0}^{t}\expo{-f\pars{t - s}}
\bracks{{\rm F}\pars{{\rm x_{a}}\pars{s}} - {\rm G}\pars{{\rm x_{b}}\pars{s}}}
\,\dd s
\end{align}

$\ds{\tt\mbox{Now, you can draw some conclusion !!!.}}$
