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 counterexample to: $\frac{1-f(x)^2}{1-x^2}\le f'(x)$ — revisited
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 $f>x$.)


