# Proving Absolute Value Inequality with fractions

I have to prove that:

$$\frac{|a+b|}{1+|a+b|} \leq \frac{|a|}{1+|a|} + \frac{|b|}{1+|b|}$$

This is where I am, trying to move from the right hand side to the other: $$\frac{|a|}{1+|a|} + \frac{|b|}{1+|b|} \geq \frac{|a|}{1+|a|+|a+b|+|b| } + \frac{|b|}{1+|b| + |a+b| + |a|} = \frac{|a|+|b|}{1+|b|+|a+b|+|a|}$$

So basically I want to keep adding to the denomerator so I can make the argument smaller, but I'm stuck.

Any hints?

Thanks!

• Hint: Use the fact that $\frac{x}{1+x}$ is monotone increasing and the fact that $\|\cdot\|$ satisfies the triangle inequality. – Jacky Chong Oct 22 '16 at 18:42
• @JackyChong imho you should post this as an answer, perhaps explain what you mean by $\|.\|$ – user159517 Oct 22 '16 at 19:03

As suggested by @user159517, I have posted a short solution.

Suppose $0\leq x \leq y$, then it follows \begin{align} x(1+y)=x+xy \leq y+xy = y(1+x) \end{align} which means \begin{align} \frac{x}{1+x} \leq \frac{y}{1+y}. \end{align} Since $|a+b| \leq |a|+|b|$, then it follows from the above monotonicity property \begin{align} \frac{|a+b|}{1+|a+b|} \leq&\ \frac{|a|+|b|}{1+|a|+|b|} = \frac{|a|}{1+|a|+|b|} + \frac{|b|}{1+|a|+|b|}\\ \leq&\ \frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}. \end{align}

If $ab>0$ we need to prove that $\frac{a+b}{1+a+b}\leq\frac{a}{1+a}+\frac{b}{1+b}$ for posittives $a$ and $b$, which gives

$2ab+ab(a+b)\geq0$, which is obvious.

If $a\geq0\geq b$ or $b\geq0\geq a$ it's enough to prove that

$\frac{a-b}{1+a-b}\leq\frac{a}{1+a}+\frac{b}{1+b}$ for non-negatives $a$ and $b$ such that $a\geq b$, which gives $ab(a-b)+2b+2ab\geq0$.

Done!