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Let $x \in (0,1)$ be arbitrary and $\epsilon >0$ as small as you wish it to be.

Now we want to find an expression for $0<f(x)<1$ such that

$$f(x)\left(1-x - (1+\epsilon) \frac{f(x)}{1-f(x)^2}\right) > 1/c$$

for some small constant $c \in \mathbb{N}$, so say $c=2$. Possibly, $c$ can also depend on $x$, which might be necessary, I am afraid. It is also possible to replace $1/c$ by $1-cx$ or $x/c$ or $x^2/c$ instead, if this helps.

Is this possible? How can one find it?

user136457
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    The equality $f(x)\left(1-x - (1+\epsilon) \frac{f(x)}{1-f(x)^2}\right) = 1/c$ is a third-degree polynomial in $f(x)$ (after multiplying by $1-f(x)^2$) and can therefore be solved exactly. Solutions of the inequality lie in between solutions of the equality. – Wouter Jul 29 '16 at 09:44
  • Thanks for the hint. This is what I also tried, but this just gives me http://www.wolframalpha.com/input/?i=solve+(y(1-x)+-y%5E3(1-x)-(1%2Bs)y%5E2%3D1%2Fc+(1-y%5E2),+y) in wolfram alpha (when setting $s:=\epsilon$ and $y:=f(x)$), and I don't really know what to do with it now... So for example do I know that for all $x \in (0,1)$ I can find a solution $f(x) \in (0,1)$ as well? – user136457 Jul 29 '16 at 10:17

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It appears that only one solution of the equality $$g(x)\left(1-x - (1+\epsilon) \frac{g(x)}{1-g(x)^2}\right) = 1/c$$ is real. Also, the lhs. changes sign at $g(x)=-1$.

There is a region in $(x,g)$-space, bounded by $g=-1$ on one side and $g=$real root of the equality on the other side, between which the inequality holds, as shown in the following plot (with $\epsilon=0,c=2$):

Region of g,x space in which inequality holds

so $f$ can be chosen anywhere in the shaded region, anywhere between the real root and -1.

Wouter
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