Solve inequality $\left|\frac{1-\lambda}{(a+1)^2}\right|<1$

Question

Let $a,\lambda \in \mathbb{R}, \lambda>0$ $$f(x)=\frac{x+\lambda}{x+1}$$ $$a=\frac{a+\lambda}{a+1}\Rightarrow a^2=\lambda\Rightarrow a=\pm\sqrt\lambda$$ $$f'(x)=\frac{1-\lambda}{(x+1)^2}$$

I want to know when $|f'(a)|<1 $, so first I calculate $|f'(a)| $ : $$|f'(a)|=\left|\frac{1-\lambda}{(a+1)^2}\right|=\left|\frac{1-a^2}{(a+1)^2}\right|=\left|\frac{1-a}{1+a}\right|$$

Then I solved the inequality $|f'(a)|<1$: $$\left|\frac{1-a}{1+a}\right|<1\Rightarrow -1<\frac{1-a}{1+a}<1\Rightarrow$$ $$0<\frac{2}{1+a}\,\hspace{0.25cm} \text{and}\hspace{0.25cm} \frac{2a}{1+a}>0 $$ $$-1<a\,\hspace{0.25cm} \text{and}\hspace{0.25cm} (-1>a \hspace{0.25 cm} \text{or} \hspace{0.25cm}a>0) $$ $$\boldsymbol{a>0} $$ So $a$ should be $a=\sqrt\lambda$ to $|f'(a)|<1$.

Is my work correct? or Is there a nother way to do it?

Answer 1

Hint: you can write $$\sqrt{|1-\lambda|}\le |a+1|$$ So you have two cases: $$a\geq -1$$ then you will get $$\sqrt{|1-\lambda|}-1\le a$$ And if $$a<-1$$ then you will get $$\sqrt{|1-\lambda|}\le -a-1$$ or $$a\le -\sqrt{|1-\lambda|}-1$$