Do there exist nonconstant real valued functions $f$ and $g$ such that the expression: $$f(x) v/g(x)$$ is maximized at $x = v$ for all positive real $v$?

Take $f(x)=(x+1)e^{x}$ and $g(x)=e^x$, then $f(x)v/g(x)=(x+1v)e^{x}$ and the derivative with respect to $x$ is $(vx)e^{x}$. 


By the arithmeticgeometric mean inequality, when $v$ is positive $$x^{\frac{1}{2}}\frac{v}{x^{\frac{1}{2}}}$$ is maximized at $x=v$ and $x=v$. 


Let $f$ be arbitrary (but nonconstant, realvalued, and differentiable), let $h$ be any antiderivative of $f'(x)/x$, and let $g=1/h$; then $f'(v)g'(v)=v$, so $fv/g$ has a critical point at $x=v$. Now you can look for conditions under which that critical point is a maximum. 


The following calculation suggests that a nice probability interpretation may exist for any solution one can construct. $u(x) = f(x)  v/g(x)$ and all its $x$ derivatives are linear functions of $v$ with coefficients that are functions of $x$. Thus, to have an extremum at $x=v$ the first derivative $u'$ must be of the form $(vx)M(x)$. Integrating the $v$degree 0 and 1 parts of this equation produces $f$ and (the reciprocal of) $g$. Algebraically this will be equivalent to Gerry's solution. The interesting points are that:
So there might be a simple probability inequality lurking behind most of the solutions. 

