<s>[INST] What is the family of functions $f(x),g(x)$ such that $f/g=x$ Suppose we have two functions $f(x)$ and $g(x)$, defined on $x\geq1$, we know
$$f(1)=g(1)=1$$
$$g,f>0$$
$$\frac{\text{d} f}{\text{d} x}\geq0,\quad\frac{\text{d} g}{\text{d} x}\leq0$$
and
$$\frac{f}{g}=x$$
for $x\geq1$.
My approach was this:
Clearly
$$f=xg$$
So we only need to define a general solution for $g$. Let us look for what exponent values, $n$, hold if we let
$$g=x^{-n}$$
giving
$$f=x^{1-n}$$
thefore we find that
$$0\geq g'=-nx^{-n-1}\Rightarrow n\geq 0$$
$$0\leq f'=(1-n)x^{-n}\Rightarrow 1\geq n$$
Therefore
$n\in[0,1]$. Are there any other possible solutions not of this form? I have tried a power series expansion, however this isn't greatly helpful as I have shown these functions behave nicely with non-integer exponents. 
 [/INST] Since $f(x)=xg(x)$, then $f'(x)=xg'(x)+g(x)$ and since $f'(x)\ge0$ then $ xg'(x)+g(x) \ge0$ which implies that $xg'(x)\ge-g(x)$......(*) 
Now since $x\ge 1$ and $g(x)\ge 0$ so that $xg(x)\ge0$, therefore from (*) we have, $\frac{g'(x)}{g(x)}\ge-\frac{1}{x} $$\Rightarrow$ $\ln g(x)\ge -\ln x $ $\Rightarrow$ $g(x) \ge\frac{1}{x}$.
Define a function $g(x)=\frac{a}{bx+c}$, where, $a=b+c$; $a,b,c>0$; $b+c\ge1$.
Then, clearly $g(x)= \frac{a}{bx+c}\ge\frac{1}{x}$, since $bx+c\ge x$ for all $x\ge 1$, we have $\frac{a}{bx+c}=\frac{b+c}{bx+c}\ge\frac{1}{x}$
$\Rightarrow $ $bx+cx \ge bx+c \Rightarrow x \ge 1$. 
Now, at $x=1$, $g(1)=\frac{a}{b+c}=1$ iff $a=b+c$. Also, $g'(x)=-\frac{ab}{(bx+c)^2}\le 0$ iff $ab>0$ 
Note:
$ab<0$ is impossible case since when $ab<0$ $\Rightarrow$ $(b+c)b<0$ but $b+c\ge 1$ so that $b<0$ in this case $bx+c\ge bx+(1-b) = b(x-1)+1$ would be greater that $x$ iff $b(x-1)>x-1$ $\Rightarrow$ $b\ge 1$ which is a contradicts the assumption that ($b<0$).
Now, since $f(x)=xg(x)$, then we define $f$ to be $f(x)=\frac{ax}{bx+c}$, where $a,b,c$ are defined above.
$$f(x)=\frac{(b+c)x}{bx+c}\Rightarrow f(1)=1.$$ Also, $f'(x) = \frac{(bx+c)a-abx}{(bx+c)^2}$ would be $\ge$ than $0$ for all $x\ge 1$ iff $(bx+c)a-abx\ge 0$ $\Rightarrow$ $abx+ac-abx\ge 0$ $\Rightarrow$$ac=(b+c)c\ge c \ge 0$, (since $b+c\ge 1$).
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