# Triangle Inequality question with fractions

Given $a, b \in \mathbb{R}$, prove that

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

When does equality hold?

The only useful thing I could get (using the triangle inequality) is:

\begin{align}1 + \left|\left(a + \frac{ab}{2}\right) + \left(b - \frac{ab}{2}\right)\right| &\le 1 + \left|a + \frac{ab}{2}\right| + \left|b - \frac{ab}{2}\right|\\ &\le 1 + |a| + |b| + |ab|\\ &=(1 + |a|)(1 + |b|)\end{align}

But it seems to be inapplicable! Any ideas?

• Try to show that if $x+y\ge z$ then, $\frac{x}{1+x}+\frac{y}{1+y}\ge \frac{z}{1+z}$. – r9m Mar 4 '14 at 12:58

Because the function $f(x)=\frac{1}{1+\frac{1}{x}}=\frac{x}{1+x}$ is increasing, we have $$\frac{|a+b|}{1+|a+b|} = \frac{1}{1+\frac{1}{|a+b|}} \le \frac{1}{1+\frac{1}{|a|+|b|}}=\frac{|a|}{1+|a|+|b|}+\frac{|b|}{1+|a|+|b|}\le\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}$$ In the first inequality we have used the triangle inequality $|a+b|\le |a|+|b|$ and in the second inequality, the positivity of $|a|$ and $|b|$.
That is, equality holds if and only if $a=0$ or $b=0$ because otherwise the last inequality is strict.
• @YiyuanLee One case is when both $a,b$ are 0 – Hawk Mar 4 '14 at 13:03
• If $a=0$ or $b=0$, otherwise the last inequality is strict. – J.R. Mar 4 '14 at 13:03