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
Best Answer
$\newcommand\les\lesssim\newcommand\R{\mathbb R}$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 $x$.