Find minimum value of a Find the minimum value of $a$ if there's a differentiable function in $\mathbb{R}$ for which :
$$e^{f'(x)}= a {\frac{|(f(x))|}{|(1+f(x)^2)|}}$$ for every $x$
pretty much stuck. I think the minimum value should be $1$ but not sure.
 A: Taking logarithms, $f'(x) = \log(a) + \log |f(x)| - \log (1 + f(x)^2)$.
This is an autonomous differential equation.  WLOG we can look at $f > 0$.  An equilibrium solution would
be a positive constant $c$ such that $\dfrac{ac}{1+c^2} = 1$.  This has real solutions iff $a \ge \ldots$.  On the other hand, if $a < \ldots$, consider what the phase-plane looks like.
A: Let $f(x)$ be any solution of the DE for any $a$ over all $\mathbb{R}$:
$$e^{f'(x)} = a \left|\frac{f(x)}{1+f(x)^2}\right|\tag{*}$$
Notice the L.H.S $> 0$ for any choice of $f(\cdot)$ and $x$. This means we must have $a > 0$ and $f(x) \ne 0$ for any $x$ and hence $f(\cdot)$ is having the same sign over all $\mathbb{R}$. Let us consider the case $f(0) > 0$ first. We have:
$$\begin{align} &1 + f'(x) \le e^{f'(x)} = a \frac{f(x)}{1+f(x)^2} = \frac{a}{f(x) + f(x)^{-1}} \le \frac{a}{2}\\
\implies & f'(x) \le \frac{a-2}{2}, \text{ for all } x \in \mathbb{R}\\
\implies & f(x) \le f(0) + \frac{a-2}{2} x, \text{ for all } x \in [0,\infty)
\end{align}$$
If $a < 2$, this leads to the contradiction that $f(x)$ changes sign before $2 f(0)/(2 - a)$. 
When $f(0) < 0$ and $a < 2$, apply the same argument above to $-f(-x)$, we see $f(x)$ changes sign somewhere between $0$ and $2 f(0)/(2-a)$ again. From these we can conclude if the DE $(*)$ has a solution, we must have $a \ge 2$. 
If $a$ is indeed $\ge 2$, it is trivial to see $(*)$ has a constant solution. 
This implies the desired minimum value of $a$ is $2$.
