[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 $p\in(0,1)$.[/step]

[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: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*}[/step]

[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: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]