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)?
 A: $$\frac{1+x}{1-x} > e^{2x}\iff \ln\left(\frac{1+x}{1-x}\right)>2x$$
$$\ln\left(\frac{1+x}{1-x}\right)=\ln(1+x)-\ln (1-x)>x-\frac{x^2}2-\left(-x-\frac{x^2}2 \right)=2x$$
A: For all $x\in\mathbb{R}$, we have
$$
e^{-2x}\ge1-2x
$$
Therefore, multiplying both sides by $e^{2x}$ gives
$$
1\ge(1-2x)\,e^{2x}
$$
Integrating from $0$ to $x$ yields
$$
x\ge(1-x)\,e^{2x}-1
$$
which, for $x\in[0,1)$, implies
$$
\frac{1+x}{1-x}\ge e^{2x}
$$
A: set $$f(x)=\ln\left(\frac{1+x}{1-x}\right)-2x$$ and use calculus
then we get $$f(0)=0$$ and $$f'(x)=-2\,{\frac {{x}^{2}}{ \left( 1+x \right)  \left( -1+x \right) }}>0$$ for $0<x<1$
A: Your method is correct. Note $f(0)=0$. You do want to show $f'(x)<0$, so that the function is decreasing.
A: Let $$f(x)=(1+x)e^{-x}-(1-x)e^x \implies f'(x)=x(e^x-e^{-x})>0, x\in(0,1)$$
So $f(x)$ is an increasing function in this domain. Then
$$f(x)\ge f(0)=0\implies \frac{1+x}{1-x}\ge e^{2x}$$
A: Hint: Apply taylor series expansion for $(1-x)^{-1}$ and $e^{2x}$
A: The question is a little bit older, but it seems people still are looking for this answer (including me, had to solve it kinda with some answers / research on my own). There is a clear way to show it, so I hope my answer will clarify this to everyone!
We have:
$$ \frac{1+x}{1-x} > e^{2x} $$
for $$ 0 < x < 1$$
The first step is to create a function $f(x)$ out of the two given terms in the inequality. One can choose if you want to show $$ f(x) = \frac{1+x}{1-x} - e^{2x} > 0 $$ or $$ f(x) = e^{2x} - \frac{1+x}{1-x} < 0 $$
It is completely up to you how you choose $f(x)$ to be, just make sure the $<$ and $>$ signs are correct. For this example, I use:
$$ f(x) = \frac{1+x}{1-x} - e^{2x} > 0 $$
Now that we have $f(x)$ defined, check the $0$ value as a lower bound:
$$ f(0) = 0$$
Okay, so let's get the first derivative of $f(x)$:
$$ f'(x) = \frac{1+x}{1-x} - e^{2x} > 0 $$
we can use $\ln$ to elimnate the euler variable:
$$ f'(x) = \ln\left(\frac{1+x}{1-x}\right) - 2x $$
by definition we can transform this into:
$$ f'(x) = \ln(1+x)-\ln(1-x) - 2x $$
$$ f'(x) = \frac{2}{(1+x)(-x+1)} -2  $$
$$ f'(x) = \frac{2x^2}{(1+x)(-x+1)} > 0 $$
and because our last term:
$$ f'(x) = \frac{2}{(1+x)(-x+1)} -2  > 0$$
is always for any value $0 < x < 1$ greater than $0$, we have proved the inequality.
Now lets take a quick look at the upper bound:
Per definition, $ 0 < x < 1 $, we already have shown that $f(x) > 0$, so let's simply plug into our function the upper bound value of $ 1 $:
$$ f'(1) = \frac{2x^2}{(1+1)(-1+1)} = 0 $$ which means that for $x = 1$, our inequality is not more true.
Therefor our inequality is true for $ 0 < x < 1 $.
This screenshot shows our $ f'(x)$ graph plotted, which cleary shows that it's always definied positive for $ 0 < x < 1 $, for any value above or under, it drops to the negative:

Hope this helps!
Greetings!
