# Inequality absolute value

If $x>0,y>0$, why does it follow for any $y \in (x,x+1)$, that from:

$$|f(y)-f(x)|<1,$$ we have $$|f(y)| \leq |f(x)|+1$$

• This is always true, no matter what $x$ and $y$ are. – Alex Becker May 3 '12 at 2:15

$|f(y)| = |(f(y)-f(x))+f(x)| \leq |f(y)-f(x)| + |f(x)| < 1 + |f(x)|$, so.
Abhishek is correct, you can also use the other triangle inequality $|f(y)-f(x)|\geq |f(y)|-|f(x)|$ This fact gives $|f(y)-f(x)|<1 \implies |f(y)|-|f(x)|<1 \implies |f(y)|<1+|f(y)|$
More generally, if $|a-b|<1$, then $|a|\le|b|+1$. Abhishek's already given a real proof using the triangle inequality, but it's also useful to be able to see what's going on in intuitive terms.
The inequality $|a-b|<1$ says that the distance between the numbers $a$ and $b$ is less than $1$. If $a$ and $b$ are on the same side of $0$, that says that the distance between $|a|$ and $|b|$ is less than $1$, so $|a|$ can't be more than $|b|+1$. (It doesn't matter which side of $0$ they're on; if you don't see this right away, draw a few sketches.)
If $a$ and $b$ are on opposite sides of $0$ and less than $1$ unit apart, $|a|$ and $|b|$ must both be less than $1$. (Again, draw a sketch or two if necessary.) Thus, $|a|<1\le |b|+1$.