Absolute Value Inequality Implication

Question

From my knowledge of the properties of absolute value inequalities, I'm failing to see why:

$$|f(x)-A|<1 \rightarrow |f(x)|<|A| + 1$$

The only approach I know is to split up the two possible cases:

$$ (1) : f(x) - A> -1$$

and

$$(2) : f(x) - A < 1$$

So $f(x) < A+1$, but how does this imply $|f(x)|<|A| + 1$ ?

Answer 1

The triangle inequality says $|a+b|\leq |a|+|b|$. From this, setting $a=(x-y)$ and $b=y$, we get $$|x| \leq |x-y|+|y|$$ from which we deduce $$|x|-|y|\leq |x-y|.$$ This inequality gives you the result you want: $$|f(x)|-|A| \leq |f(x)-A| \lt 1$$ so from $|f(x)|-|A| \lt 1$ you immediatly get $$|f(x)|\lt |A|+1.$$ In fact, the following holds: try to verify it: $$\Bigl| |x|-|y|\Bigr| \leq |x\pm y| \leq |x|+|y|.$$

Answer 2

More generally: This is a consequence of the so-called reverse triangle inequality, itself a consequence of the triangle inequality: $$ ||x|-|y|| \leq |x-y|, \qquad x,y\in\mathbb{R} \tag{$\dagger$} $$ Apply it to $f(x)$ and $A$ to get $$ ||f(x)|-|A|| \leq |f(x)-A| $$ which implies $|f(x)|\leq |f(A)|+|f(x)-A|$.

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The reverse triangle inequality, very useful, is quite immediate given the triangle inequality: from $$ |x| = |x-y+y| \leq |x-y|+|y| $$ and $$ |y| = |y-x+x| \leq |x-y|+|x| $$ we get $||x|-|y||= \max( |x|-|y|, |y|-|x| ) \leq |x-y|$.