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Christophe Leuridan
Christophe Leuridan
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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 $$f(x) := \frac{1}{2} [x+1-(1-x)^3]/2.$$$$f(x) := \frac{1}{2} [x+1-(1-x)^3].$$ 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 $1$.

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 $$f(x) := \frac{1}{2} [x+1-(1-x)^3]/2.$$ 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 $1$.

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 $1$.

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Christophe Leuridan
Christophe Leuridan
  • 3.8k
  • 6
  • 15

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 $$f(x) := \frac{1}{2} [x+1-(1-x)^3]/2.$$ 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 $1$.