$\frac{1+\sqrt x}{1-\sqrt x} \leq e^{Cx}\frac{1+x}{1-x}$ I try find a constant $C$ such that
$$\frac{1+\sqrt x}{1-\sqrt x} \leq e^{Cx}\frac{1+x}{1-x}$$
Maybe is not posible
 A: $$\frac{1+\sqrt x}{1-\sqrt x} \leq e^{Cx}\frac{1+x}{1-x}$$
can be simplified as
$$1+\frac{2\sqrt x}{1+x} \leq e^{Cx}.$$
The LHS equals $1$ and has an infinite derivative at $x=0$.
A: You are right, it is not possible.
$$ f(x)=\log\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\cdot\frac{1-x}{1+x}\right) $$
is a concave function in a right neighbourhood of the origin (just consider the Taylor series of $f(x^2)$), but $\lim_{x\to 0^+}f'(x)=+\infty$, hence it is not possible to find a constant $C$ such that $f(x)\leq C x$ holds in a right neighbourhood of the origin.
A: Let $\sqrt{x}=t$, so we have
$$
\frac{1+t}{1-t}\le e^{Ct^2}\frac{1+t^2}{1-t^2}
$$
(for $t\ge0$ and $t\ne1$). For $t>1$ we can multiply this by $1-t^2<0$ getting
$$
\frac{(1+t)^2}{1+t^2}\ge e^{Ct^2}
$$
that forces $C\le0$, by taking the limit at $\infty$.
For $0\le t<1$ we get instead
$$
\frac{(1+t)^2}{1+t^2}\le e^{Ct^2}
$$
and, by continuity, evaluating at $1$,
$$
2\le e^C
$$
that gives $C\ge\ln2>0$. A contradiction.
If you allow $e^{Cx+D}$, you have again $C<0$ and $C+D\ge\log2$, then any value $D>\log2$ and $C=-D+\log2$ will do.
