Solve the inequality $\left|\frac{1-a}{1+a}\right|<1$

Question

Are these methods correct? Which one is better?

First Method: $$\left|\frac{1-a}{1+a}\right|<1\Rightarrow |1-a|<|1+a|\Rightarrow(1-a)^2<(1+a)^2\Rightarrow -2a< 2a\Rightarrow 4a>0\Rightarrow \boldsymbol{a>0}$$

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Second Method

$$\left|\frac{1-a}{1+a}\right|<1$$ $$-1<\frac{1-a}{1+a}<1\Rightarrow$$ $$0<\frac{2}{1+a}\,\hspace{0.25cm} \text{and}\hspace{0.25cm} \frac{2a}{1+a}>0 $$ $$a>-1\,\hspace{0.25cm} \text{and}\hspace{0.25cm} (a<-1 \hspace{0.25 cm} \text{or} \hspace{0.25cm}a>0) $$ $$\boldsymbol{a>0}$$

Answer 1

Since none of oldest two answers addresses OP's two prompts at the beginning of the post, I post another answer for fun.

Are these methods correct?

Yes, definitely because each implication is irrefutable. However, they are incomplete because one has to find all $a$ satisfying the inequality. OP has shown that

$$\left|\frac{1-a}{1+a}\right|<1 \implies a > 0,$$

but does each $a>0$ satisfy this inequality? --- This is the missing part of OP's attempt.

To complete the solution, we prove the converse of the above implication.

• $a \ge 1$: $\left|\dfrac{1-a}{1+a}\right| = \dfrac{a-1}{1+a} = 1 - \dfrac{2}{1+a} < 1$
• $0 < a < 1$: $\left|\dfrac{1-a}{1+a}\right| = \dfrac{1-a}{1+a}$. The numerator $1-a \in (0,1)$ and the denominator $1+a > 1$, so $\dfrac{1-a}{1+a} < 1$.

Which one is better?

The first one is more concise, and I personally prefer this one in this case. However, the second one is more flexible and informative: first, it gives more information about the positivity of the fraction $\dfrac{1-a}{1+a}$ itself by decomposing it into several cases. Second, the first method (squaring and compare the two quadratic polynomials in $a$) works well if

• either the coefficient of $a$ in both the numerator and the denominator are equal; or
• the difference of squares of both the numerator and the denominator can be factorized (in $\Bbb{R}$, by solving roots).

Otherwise, I think the first method would not save you work.

Answer 2

Third method

$|1-z|<|1+z|$

Set $A=(-1,0)$ and $B=(1,0)$ and $M=(x,y)$ with $z=x+iy$

You search $M$ in the plane such that $BM<AM$, these are the points in the half-plane delimited by the perpendicular bissector of $[AB]$ (i.e. $x=0$) and closer to $B$, namely $x>0$.

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Note: for the real case only

Just put $A=-1$ and $B=+1$ on the real line and you search for $M$ such that $BM<AM$, so $M$ is on the right of $I=0$ the middle of $[AB]$.

The condition is thus $a>0$.

Answer 3

I would always suggest sketching the graphs.

Draw the graphs $y=|1-x|$ and $y=|1+x|$ and let the graph do the rest.

Answer 4

Both methods are OK. I would vote for the 1.st one. It is shorther.

You could do also like this:

$$ |{2\over 1+a}-1|<1$$ so $$ -1<{2\over 1+a}-1<1$$ so $$ 0<{2\over 1+a}<2$$

From the left inequality we deudce that $1+a>0$ so we have $2< 2a+2$ or $a>0$.