Assume that $f:[0,1]\to [0,1]$ is a diffeomorphism so that $(f''(x)/f'(x))'<0$ and that $f''(0)=0$. It seems to me that $$\frac{1-f(x)^2}{1-x^2}\le f'(x)$$ on $[0,1]$. But no proof so far.
A condition on the inequality $f'(x)/(1-f(x)^2)-1/(1-x^2)\ge 0$
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