Show that : $$x < \log\Bigl(\frac{1}{1-x}\Bigr) < \frac{x}{1 - x}\,;$$ If $$0 < x < 1$$
Solution:
If $$f(x) = \log\Bigl(\frac{1}{1 - x}\Bigr)$$ Then the function is continuous in [0, x] And also differentiable in (0, x) So We can apply Lagrange's MVT on f(x).
Fine!
So $$f'(x) =\frac{1}{1 - x}$$ As per LMVT : $$f(x) - f(0) = xf'(θx) ;$$ Where : $$0 < θ < 1$$
=> $$xf'(θx) = - \log(1 - x) $$
=> $$\frac{x} {1 - θx} = - \log(1 - x)$$
=> $$\frac{x} {1 - θx} = - \log(1 - x)......... (1)$$
So, $$0 < θx < x$$
Or $$0 > - θx > - x $$
Or $$1 > 1 - θx > 1 - x $$
Or $$1 < \frac{1} {1 - θx} < \frac{1} {1 - x} $$
Or $$x < \frac{x} {1 - θx} <\frac{x}{1 - x} $$
Or $$x < - \log(1 - x) < \frac{x} {1 - x}... \text{ from } (1)$$
Or
$$x < \log\Bigl(\frac{1}{1 - x}\Bigr) <\frac{x}{1 - x}$$
Hint:
The inequalities should be reversed.
Start from the well-known inequality $$\log u\le u-1\quad\text{for all }u>0\qquad(< \text{ if } u\ne1),$$ and make the relevant substitutions.
Some details:
First set $\;u=\dfrac1{1-x}$ (which is positive since $0<x<1$). You obtain instantly $$\log\frac1{1-x} <\frac1{1-x}-1=\frac x{1-x}.$$
Now set $u=1-x$: you get $$\log(1-x)<(1-x)-1=-x\iff \log\frac1{1-x}>-(-x)=x.$$