Inequality: $\frac{1+x}{1-x} > e^{2x}$ for $0 < x < 1$ I was asked to prove the inequality $\frac{1+x}{1-x} > e^{2x}$ for $0 < x < 1$
My approach would be with the differential equation.
$f(x) = e^{2x} - \frac{1+x}{1-x} $
$f(x)' = 2e^{2x} − \frac{x+1}{(1−x)^2} − \frac{1}{1−x} = 2e^{2x}−\frac{2}{(1−x)^2} $
Now it would suffice to say that $f(x)' > 0$ for $0 < x < 1$, but $f(x)' < 0$ for $0 < x < 1$
Could you please tell me how to solve this inequality correctly (with the differential equation)?

$$\frac{1+x}{1-x} > e^{2x}\iff \ln\left(\frac{1+x}{1-x}\right)>2x$$
$$\ln\left(\frac{1+x}{1-x}\right)=\ln(1+x)-\ln (1-x)>x-\frac{x^2}2-\left(-x-\frac{x^2}2 \right)=2x$$
