Proving or disproving that an inequality implies another inequality.

Question

I am wondering if $|a| > |b|$ implies $|\frac{b+b^{2}}{a+a^{2}}| < 1$, where $a$ and $b$ are real numbers. I have tested numerically with many cases and I have found this to be true in all of my testing.

We know that $|\frac{b+b^{2}}{a+a^{2}}| < 1$ is equivalent to saying $|a + a^{2}| > |b + b^{2}|$, and I believe this to be always true if $|a| > |b|$, but I am not sure how I would prove this easily.

One might be able to do a proof by cases, by checking each case $a > 1, 1 \geq a > 0, 0 > a \geq -1, -1 > a$ with each case $b \geq 1, 1 > b \geq 0, 0 \geq b > -1, -1 \geq b$, but this is tedious and there is probably a more elegant way (Possibly using the triangle inequality in some way?).

I appreciate any response.

EDIT: I have asked another question directly related to this one here. Once again I would greatly appreciate responses to this question as well.

EDIT 2: I have asked a second question directly related to this one here. This one, we have the added restriction that $a > 0$. I would greatly appreciate responses to this question as well.

Answer 1

Counter-example: Take $$ a_n = -\frac{1}{n} $$ and $$ b_n = \frac{1}{n} - \frac{1}{n^{2}}. $$ Then for all $n$ one has $\lvert b_n \rvert < \lvert a_n\rvert$, but $$ \left\lvert \frac{b_n+b_n^2}{a_n+a_n^2} \right\rvert = \frac{\frac{1}{n} - \frac{2}{n^{3}} + o\left(\frac{1}{n^{3}}\right)}{\frac{1}{n} - \frac{1}{n^2}} = 1+\frac{1}{n} + o\left(\frac{1}{n}\right) > 1 $$ the inequality holding for $n$ big enough.

Answer 2

You have $|b|/|a| < 1$ ($a\neq 0$). Hence $b^2 < a^2$ and thus $1+b^2 < 1+a^2$. Equivalently, $$ \frac{1+b^2}{1+a^2} < 1. $$ Now, this implies $$ \left|\frac{b+b^3}{a+a^3}\right| = \frac{|b|}{|a|}\frac{1+b^2}{1+a^2} < 1, $$ since both factors are $< 1$. So, your result is false, but it holds when you replace 2 by 3 in the exponent.

Answer 3

Take $a=-0.6$ and $b=-0.5$. Then $|a|=0.6 > 0.5 = |b|$ but $|a+a^2|=0.24 < 0.25=|b+b^2|$.

Looking at the graph of $y=|x+x^2|$ below, we see that it is increasing for $x\in [-1,-0.5]$ when $x<0$. Hence the counter-example.