A counterexample to: $\frac{1-f(x)^2}{1-x^2}\le f'(x)$ — revisited Can we find a counterexample to the following assertion?
Assume that $f:[-1,1]\to [-1,1]$ an odd function of class $C^3$, and assume thaht $f$ is a concave increasing diffeomorphism of $[0,1]$ onto itself. Then my examples say that $$\frac{1-f(x)^2}{1-x^2}\le f'(x),\; x\in (0,1).$$
 A: A counterexample is provided by any function that equals $f(x)=1+m(x-1)$ near $x=1$, with $0<m<1$. (Maybe this is in fact just restating Anthony's comment, with a typo corrected?)
What is actually true is the trivial observation that (by the mean value theorem)
$$
\frac{1-f^2(x)}{1-x^2}=f'(c) \frac{1+f(x)}{1+x} \le f'(x)\frac{1+f(x)}{1+x} ,
$$
and it seems this is as far as we can go in general. (In my example, we have equality here, and $(1+f)/(1+x)>1$, since $f>x$.)
A: I complete the reformulation given by Andrea Marino and give another counterexample.
First, the inequality of the beginning can be written
$$\forall x \in (0,1), \quad \frac{f'(x)}{1-f^2(x)} \ge \frac{1}{1-f^2(x)}$$
and means that the function $x \mapsto \arg\tanh(f(x))-\arg\tanh(x)$ is non-decreasing on $(0,1)$, or equivalently (composing with $\tanh$) that the function
$$g : x \mapsto \frac{f(x)-x}{1-xf(x)}$$
is non-decreasing on $(0,1)$.
Now, let
$$f(x) := \frac{1}{2} [x+1-(1-x)^3].$$
The function $f$ thus defined satisfies the assumptions. Let us compute the corresponding function $g$.
\begin{eqnarray*}
g(x) &=& \frac{[x+1-(1-x)^3]-2x}{2-x[x+1-(1-x)^3]} \\
&=& \frac{1-x-(1-x)^3}{2-x-x^2+x(1-x)^3} \\
&=& \frac{1-(1-x)^2}{2+x+x(1-x)^2} \\
&=& \frac{2x-x^2}{2+2x-2x^2+x^3} \\
\end{eqnarray*}
The quantities $u(x):=2x-x^2$ and $v(x):=2+2x-2x^2+x^3$ are positive on $[0,1]$ (since $x \ge x^2$ on $[0,1]$), and $u'(1)=2-2=0$ whereas $v'(1)=2-4+3=1$. Hence $g$ is decreasing at the neighborhood of $1$.
