Given $\left|a\right|<1$ and $\left|b\right|<1$, prove that $\left|\frac{a+b}{1+ab}\right|<1$.

Question

Here is my question:

Let $a$ and $b$ be real numbers satisfying $\left|a\right|<1$ and $\left|b\right|<1$, prove that $\left|\frac{a+b}{1+ab}\right|<1$.

Her is my attempt:

• It is easy to note that $1+ab>0$.
• We can rewrite the inequality into $-1<\frac{a+b}{1+ab}<1$.

From here I can’t see the next step to prove it. Any suggestions?

Answer 1

we have to prove $$|a+b|<|1+ab|$$

or $$a^2+b^2+2ab< 1+a^2b^2+2ab$$

or $$(a^2-1)(1-b^2)< 0$$ which is true because $|a|<1$,$|b|<1$

Answer 2

We know that $-1<a<1$ and $-1<b<1$.

Therefore,

• $(1-a)(1-b)>0\Longrightarrow (1+ab)-(a+b)>0\Longrightarrow 1+ab>a+b$
• $(1+a)(1+b)>0\Longrightarrow (1+ab)+(a+b)>0\Longrightarrow 1+ab>-(a+b)$

We get $$\left|1+ab\right|=1+ab>\left|a+b\right|$$, which is equivalent to $$\left|\frac{a+b}{1+ab}\right|<1$$ QED.

Answer 3

Because $|a|,|b| < 1$, then there exists $u,v$ such that $a=\tanh(u)$ and $b=\tanh(v)$. Then $$\left| \frac{a+b}{1+ab}\right| = \left| \frac{\tanh(u)+\tanh(v)}{1+\tanh(u)\tanh(v)}\right| = \left|\tanh \left(u+v \right)\right| <1$$

Answer 4

We have that

$$\left|\frac{a+b}{1+ab}\right|<1 \iff -1-ab\le a+b \le1+ab$$

and

$$-1-ab\le a+b \iff a+ab+b +1 \ge 0 \iff (a+1)(b+1) \ge 0$$

$$ a+b \le1+ab \iff a-ab+b-1\le 0 \iff (a-1)(b-1) \le 0$$