Absolute Value. Real Analysis

Question

I just came across the following.

The working goes from this step: $|f(x)-f(c)|<1$.

To the next step: $|f(x)|<1+|f(c)|$

Could someone please explain the transition from one step to the next using absolute value properties.

Thanks!

Answer 1

Hint

The complete triangle inequality is

$$|a|-|b|\le|a\pm b|\le|a|+|b|$$

Answer 2

Actually the inequality of interest can be more general, and, surprisingly, such an inequality can be easily derived from the fundamental triangle inequality, i.e. an inequality of the form $|a+b| \leq |a| + |b|$.

It holds true that $||a|-|b|| \leq |a-b|$ for all reals $a,b$. (Your inequality follows from that $|a| - |b| \leq ||a|-|b||$.) Let $a,b$ be real. Note that $|a| = |b+a-b| \leq |b| + |a-b|$ by triangle inequality; note that $|b| = |a-b -a| \leq |a-b| + |a|$ by triangle inequality. So $|a|-|b| \leq |a-b|$ by the first observation, and $|a|-|b| \geq -|a-b|$ by the second observation. Combining these results we have $||a| - |b|| \leq |a-b|$.

As a side remark possibly of your potential interest, note that this result is true in any metric space. (We have only employed triangle inequality, which is an axiom for any metric space.)