[proofplan]
We prove the two implications separately. If $F\ge G$ pointwise, then every point where $G$ has reached level $p$ is also a point where $F$ has reached level $p$, so the level set defining the quantile of $F$ contains the corresponding level set for $G$, forcing the infimum for $F$ to be smaller. Conversely, if the quantile inequality holds and $F(x)<G(x)$ at some point, we choose an intermediate level $p$ and use the inverse inequality for quantiles to obtain $G^{-1}(p)\le x<F^{-1}(p)$, contradicting the assumed quantile order.
[/proofplan]
[step:Show that pointwise domination of distribution functions reverses the level-set infima]
Assume that
\begin{align*}
F(x)\ge G(x)\quad\text{for every }x\in\mathbb R.
\end{align*}
Fix $p\in(0,1)$. Define the level sets $A_F(p)\subset\mathbb R$ and $A_G(p)\subset\mathbb R$ by
\begin{align*}
A_F(p):=\{x\in\mathbb R:F(x)\ge p\}
\end{align*}
and
\begin{align*}
A_G(p):=\{x\in\mathbb R:G(x)\ge p\}.
\end{align*}
Because $F$ and $G$ are distribution functions and $p\in(0,1)$, both sets are nonempty and bounded below: nonemptiness follows from the limit $F(x),G(x)\to1$ as $x\to\infty$, and boundedness below follows from the limit $F(x),G(x)\to0$ as $x\to-\infty$ together with monotonicity.
If $x\in A_G(p)$, then $G(x)\ge p$, and the pointwise hypothesis gives $F(x)\ge G(x)\ge p$. Hence $x\in A_F(p)$, so
\begin{align*}
A_G(p)\subset A_F(p).
\end{align*}
Since both sets are nonempty and bounded below, inclusion gives
\begin{align*}
\inf A_F(p)\le \inf A_G(p).
\end{align*}
By the definitions of the quantile functions, this is
\begin{align*}
F^{-1}(p)\le G^{-1}(p).
\end{align*}
Since $p\in(0,1)$ was arbitrary, the quantile inequality holds for every $p\in(0,1)$.
[guided]
Assume that $F(x)\ge G(x)$ for every $x\in\mathbb R$. We want to compare the two quantiles at a fixed probability level, so fix $p\in(0,1)$ and introduce the two threshold sets
\begin{align*}
A_F(p):=\{x\in\mathbb R:F(x)\ge p\}
\end{align*}
and
\begin{align*}
A_G(p):=\{x\in\mathbb R:G(x)\ge p\}.
\end{align*}
These are exactly the sets whose infima define $F^{-1}(p)$ and $G^{-1}(p)$.
We first check that the infima are legitimate real quantities. Since $F$ and $G$ are distribution functions, each is nondecreasing, has limit $0$ at $-\infty$, and has limit $1$ at $+\infty$. Because $p<1$, each function eventually reaches level $p$, so $A_F(p)$ and $A_G(p)$ are nonempty. Because $p>0$, each function is eventually below $p$ on the far left, and monotonicity then prevents any sufficiently far-left point from belonging to the corresponding level set. Thus both $A_F(p)$ and $A_G(p)$ are bounded below.
Now take any $x\in A_G(p)$. By definition of $A_G(p)$, this means $G(x)\ge p$. The pointwise order says $F(x)\ge G(x)$ at the same point $x$, so
\begin{align*}
F(x)\ge G(x)\ge p.
\end{align*}
Therefore $x\in A_F(p)$. Since every element of $A_G(p)$ is also an element of $A_F(p)$, we have
\begin{align*}
A_G(p)\subset A_F(p).
\end{align*}
For nonempty subsets of $\mathbb R$ that are bounded below, passing to a larger set can only move the infimum leftward or leave it unchanged. Hence
\begin{align*}
\inf A_F(p)\le \inf A_G(p).
\end{align*}
Using the definitions of the two quantile functions, this becomes
\begin{align*}
F^{-1}(p)\le G^{-1}(p).
\end{align*}
Because no special property of $p$ was used except $p\in(0,1)$, this proves the quantile inequality for every $p\in(0,1)$.
[/guided]
[/step]
[step:Use an intermediate level to recover pointwise domination from quantile domination]
Assume that
\begin{align*}
F^{-1}(p)\le G^{-1}(p)\quad\text{for every }p\in(0,1).
\end{align*}
We prove the contrapositive. Suppose there exists $x_0\in\mathbb R$ such that
\begin{align*}
F(x_0)<G(x_0).
\end{align*}
Since $F(x_0),G(x_0)\in[0,1]$, the density of $\mathbb R$ gives a number $p\in(0,1)$ such that
\begin{align*}
F(x_0)<p<G(x_0).
\end{align*}
Apply [citetheorem:8346] to the distribution function $G$, the level $p\in(0,1)$, and the point $x_0\in\mathbb R$. Since $p\le G(x_0)$, we obtain
\begin{align*}
G^{-1}(p)\le x_0.
\end{align*}
Apply [citetheorem:8346] to the distribution function $F$, the same level $p$, and the same point $x_0$. Since $F(x_0)<p$, the equivalence $F^{-1}(p)\le x_0\iff p\le F(x_0)$ implies
\begin{align*}
x_0<F^{-1}(p).
\end{align*}
Combining the two inequalities gives
\begin{align*}
G^{-1}(p)<F^{-1}(p),
\end{align*}
contradicting the assumed inequality $F^{-1}(p)\le G^{-1}(p)$. Therefore no such $x_0$ exists, and hence
\begin{align*}
F(x)\ge G(x)\quad\text{for every }x\in\mathbb R.
\end{align*}
[guided]
Assume that the quantiles are ordered:
\begin{align*}
F^{-1}(p)\le G^{-1}(p)\quad\text{for every }p\in(0,1).
\end{align*}
To prove the pointwise inequality $F\ge G$, we argue by contrapositive. Suppose that the desired pointwise inequality fails. Then there is a point $x_0\in\mathbb R$ such that
\begin{align*}
F(x_0)<G(x_0).
\end{align*}
The purpose of choosing a level strictly between these two values is to force the two quantiles to lie on opposite sides of $x_0$. Since $F(x_0)$ and $G(x_0)$ both belong to $[0,1]$, and the inequality is strict, there exists $p\in(0,1)$ satisfying
\begin{align*}
F(x_0)<p<G(x_0).
\end{align*}
Now we translate the two inequalities involving $F(x_0)$ and $G(x_0)$ into inequalities involving quantiles.
First apply [citetheorem:8346] to the distribution function $G:\mathbb R\to[0,1]$, the level $p\in(0,1)$, and the point $x_0\in\mathbb R$. Its hypotheses are exactly met: $G$ is a distribution function, $p$ lies in $(0,1)$, and $x_0$ is real. Since $p<G(x_0)$, in particular $p\le G(x_0)$, the inverse inequality gives
\begin{align*}
G^{-1}(p)\le x_0.
\end{align*}
Second apply [citetheorem:8346] to the distribution function $F:\mathbb R\to[0,1]$, again with the level $p\in(0,1)$ and the point $x_0\in\mathbb R$. The same hypotheses are satisfied. The equivalence in that theorem says
\begin{align*}
F^{-1}(p)\le x_0\quad\Longleftrightarrow\quad p\le F(x_0).
\end{align*}
But our choice of $p$ gives $F(x_0)<p$, so the right-hand condition is false. Therefore the left-hand condition is false, and hence
\begin{align*}
x_0<F^{-1}(p).
\end{align*}
Putting the two conclusions together yields
\begin{align*}
G^{-1}(p)\le x_0<F^{-1}(p).
\end{align*}
In particular,
\begin{align*}
G^{-1}(p)<F^{-1}(p).
\end{align*}
This contradicts the assumed quantile order at the same level $p$, namely $F^{-1}(p)\le G^{-1}(p)$. The contradiction shows that there is no point $x_0\in\mathbb R$ with $F(x_0)<G(x_0)$. Therefore
\begin{align*}
F(x)\ge G(x)\quad\text{for every }x\in\mathbb R.
\end{align*}
[/guided]
[/step]
[step:Combine the two implications]
The first step proves that
\begin{align*}
F(x)\ge G(x)\quad\text{for every }x\in\mathbb R
\end{align*}
implies
\begin{align*}
F^{-1}(p)\le G^{-1}(p)\quad\text{for every }p\in(0,1).
\end{align*}
The second step proves the converse implication. Hence the two stated conditions are equivalent.
[/step]
