Let $f,g \in C(\mathbb R)$ with $\exists M \in \mathbb R^*, \forall (x,y) \in \mathbb R^2, M\times (f(x)-f(y))(g(x)-g(y)) \geq 0$.
Is it true that exists $ u$ any real function, and $a,b$ monotone function with $f(x)=a(u(x))$ and $g(x)=b(u(x))$ ?
No answer here : https://artofproblemsolving.com/community/c7h3261085
About the inequality : https://fr.m.wikipedia.org/wiki/In%C3%A9galit%C3%A9_de_Tchebychev_pour_les_sommes
$\newcommand\les\lesssim\newcommand\R{\mathbb R}$The answer is yes.
Indeed, without loss of generality, we have the co-monotonicity condition $$(f(x)-f(y))(g(x)-g(y))\ge0 \tag{$-10$}\label{-10}.$$ (Otherwise, replace $g$ by $-g$.)
For real $x$ and $y$, write $y\les x$ if $f(y)\le f(x)$ and $g(y)\le g(x)$. For real $x$, let $$E_x:=\{y\in\R\colon y\les x\}.$$ The relation $\les$ is transitive, so that for all real $x$ and $y$ we have $$y\les x\implies E_y\subseteq E_x.$$
For real $x$, let now $$u(x):=\mu(E_x),$$ where $\mu$ is any finite measure whose support set is the entire real line $\R$.
It is enough to show that for all real $x$ and $y$ we have $$u(y)\le u(x)\implies f(y)\le f(x)\ \&\ g(y)\le g(x); \tag{0}\label{0}$$ see details on this at the end of this answer.
Suppose the contrary. Then without loss of generality for some real $x$ and $y$ we have $$u(y)\le u(x)\ \&\ f(y)>f(x) \tag{10}\label{10}$$ and hence, by the co-monotonicity \eqref{-10} of $f$ and $g$ we have $g(y)\ge g(x)$, so that $x\les y$ and hence $$E_x\le E_y. \tag{20}\label{20}$$
Moreover, by \eqref{10} and the continuity of $f$, there is some some $c$ between $x$ and $y$ such that $f(y)>f(c)>f(x)$ and hence $f(y)>f(z)>f(x)$ for all $z$ in some neighborhood $N_c$ of $c$. So, again by the co-monotonicity \eqref{-10} of $f$ and $g$ we have $g(y)\ge g(z)\ge g(x)$ for $z\in N_c$, so that $N_c\subseteq E_y\setminus E_x$.
So, in view of \eqref{20}, $$u(x)=\mu(E_x)\le\mu(E_y)-\mu(N_c)=u(y)-\mu(N_c)<u(y),$$ which contradicts \eqref{10}. $\quad\Box$
Details on \eqref{0}: Take any $t\in u(\R)$, so that $t=u(x)$ for some real $x$, and let $$a(t):=f(x). \tag{30}\label{30}$$ This yields a well-defined function $a\colon u(\R)\to\R$, because, in view of \eqref{0}, if $u(y)=u(x)$ for some real $x$ and $y$, then $f(y)=f(x)$. Moreover, we have $f(x)=a(u(x))$ for all real $x$. Furthermore, for any $s$ and $t$ in $u(\R)$ such that $s<t$ we have some real $x$ and $y$ such that $u(x)=s<t=u(y)$, so that, in view of \eqref{30} and \eqref{0}, $a(s)=f(x)\le f(y)=a(t)$. So, the function $a$ on $u(\R)$ is nondecreasing. If desired, the function $a$ can be extended from $u(\R)$ to $\R$ by the formula $$\tilde a(t):=\sup\{a(s)\colon s\in u(\R)\cap(-\infty,t]\}$$ for all real $t$, so that $\tilde a(t)=a(t)$ for $t\in u(\R)$. Re-denoting $\tilde a$ as $a$, we will have a nondecreasing function $a\colon\R\to\R$ such that $f(x)=a(u(x))$ for all real $x$.
Similarly, we will have a nondecreasing function $b\colon\R\to\R$ such that $g(x)=b(u(x))$ for all real $x$.
Here is a slightly more elementary proof; note that continuity of $f$ and $g$ are in fact not required.
Without loss of generality (replace $g$ by $-g$ if necessary) we can assume that $f$ and $g$ are such that $$ (f(x) - f(y))(g(x) - g(y)) \geq 0, \quad \forall x,y\in \mathbb{R}. \tag{A}\label{A}$$
Let $U\subseteq \mathbb{R}^2$ be the set $$ U:= \{(f(x), g(x)) : x\in \mathbb{R} \}$$ then expression \eqref{A} is equivalently written as $$ (s_1, t_1), (s_2,t_2)\in U \implies (s_1 - s_2)(t_1 - t_2) \geq 0. \tag{B}\label{B}$$ Geometrically this means that the vector $(s_1 - s_2, t_1 - t_2)$ must live in the first or third quadrants, or their closure. So the mental image of $U$ should be a set that travels diagonally up $\mathbb{R}^2$ always in a direction between North and East. This suggests the following lemma.
Lemma 1 The function $\phi:U\to \mathbb{R}$ given by $\phi(s,t) = s+t$ is injective.
Proof: Suppose not, then there exists $(s_1,t_1) \neq (s_2,t_2)$ such that $s_1 + t_1 = s_2 + t_2$, which implies $s_1 - s_2 = t_2 - t_1$. Since the two pairs are not equal, neither difference is zero, and this would force $(s_1 - s_2)(t_1 - t_2) < 0$, contradicting \eqref{B}. Q.E.D.
Since $\phi$ is injective, there exists $\psi:\phi(U) \to U$ such that $\phi\circ \psi$ is the identity. I claim the following:
The desired result can be obtained by letting $u(x) = f(x) + g(x)$ and $a, b$ be the first and second components of $\psi$.
Our argument above already shows that $a,b$ are well-defined, and that $a(u(x)) = f(x)$ and $b(u(x)) = g(x)$. It suffices to show that they are monotone. This is established by the following lemma.
Lemma 2 If $\phi(s,t) < \phi(s',t')$, then $s \leq s'$ and $t \leq t'$.
Proof: Assume for contradiction and WLOG that $s > s'$, then as we hypothesized $s+ t < s' + t'$ we must also have $t < t'$. But then $(s-s')(t-t') < 0$, contradicting (B). Q.E.D.
What we gain if $f$ and $g$ are assumed continuous is that $u$ is now a continuous function, and hence its image is an open interval in $\mathbb{R}$. One can show then the set $U$ is in fact a continuous curve in $\mathbb{R}^2$.
Choose an arbitrary $x_0 \in \mathbb{R}$ and define $h: \mathbb{R} \to \mathbb{R}$ by $h(x) = (f(x) - f(x_0))(g(x) - g(x_0))$. The continuity of $f$ and $g$ implies the continuity of $h$. Additionally, the condition $$(f(x) - f(y))(g(x) - g(y)) > 0$$ means that $f$ and $g$ must exhibit the same monotonic behavior (either both increasing or both decreasing) over any interval where they are monotonic. Moreover, $h(x) > 0$ for all $x \neq x_0$, showing that neither $f$ nor $g$ can be constant over any interval.
Define $u: \mathbb{R} \to \mathbb{R}$ by $$u(x) = \int_{x_0}^{x} \sqrt{|h(t)|} \, dt$$ Since $h$ is continuous, so is $|h|$ and $u$ is continuous. By the Fundamental Theorem of Calculus, $u'(x) = \sqrt{|h(x)|} $ for $x \ne x_0$. Since $h(x) > 0$ for $x \ne x_0$, we have $u'(x) > 0$ for $x \neq x_0$. To ensure $u$ is strictly increasing across its entire domain, we note that the continuity of $u$ and the positivity of $u'$ for all $x \neq x_0$ directly guarantees that $u$ must be strictly increasing over all of $\mathbb{R}$.
To ensure $u$ is invertible, a sufficient condition is that $u(x)$ becomes unbounded as $x$ approaches either positive or negative infinity, along with the requirement that $ u$ covers the entire range of $\mathbb{R}$. These conditions guarantee that for every $c \in \mathbb{R}$ there exists at least one $x$ such that $u(x) = c$. Since $h(x) > 0$ for all $x \neq x_0$, a sufficient condition for the unboundedness of $u(x)$ is that the limit $\lim_{x \to \infty} h(x)$ or $\lim_{x \to -\infty} h(x)$ exists and is non-zero.
Because $u$ is continuous and strictly increasing, the Inverse Function Theorem guarantees that its inverse function $u^{-1}: \mathbb{R} \to \mathbb{R}$ exists and is also continuous and strictly increasing. Additionally, since $u'(x) \neq 0 $ for $x \neq x_0$, the Inverse Function Theorem ensures that $u^{-1}$ is continuously differentiable on its domain. Let $a, b: \mathbb{R} \to \mathbb{R}$ be defined by $a(y) = f(u^{-1}(y))$ and $b(y) = g(u^{-1}(y))$. The continuity of $a$ and $b$ follows directly from the continuity of $f$, $g$, and $u^{-1}$. Furthermore, where $f$ and $g$ are differentiable, their compositions with the continuously differentiable $u^{-1}$ ensure that $a$ and $b$ are also differentiable.
To show that $a$ and $b$ inherit the monotonic properties of $f$ and $g$, consider any $y_1 < y_2$. Since $u$ is strictly increasing, $u^{-1}(y_1) < u^{-1}(y_2)$. Because $h(x) > 0$ for all $x \neq x_0$, it's guaranteed that $f$ and $g$ (and consequently $a$ and $b$) maintain the same monotonic behavior relative to their values at $x_0$. Thus, $a$ and $b$ will also be strictly monotone if $f$ and $g$ are strictly monotone.
Finally, note that $$u(x_0) = \int_{x_0}^{x_0}\sqrt{|h(t)|}dt = 0$$ Therefore, $a(u(x_0)) = f(u^{-1}(u(x_0))) = f(x_0)$, and similarly, $b(u(x_0)) = g(x_0))$. Since $a$, $b$, $f$, and $g$ all exhibit the same monotonic behavior relative to $x_0$, the desired representations $f(x) = a(u(x))$ and $g(x) = b(u(x))$ hold for all $x \in \mathbb{R}$.
Maybe this can be extended to other domains besides real numbers. Or even to higher dimensions.
