Prove that if $a,b \in \mathbb{R}$ and $|a-b|\lt 5$, then $|b|\lt|a|+5.$ I'm trying to prove that if $a,b \in \mathbb{R}$ and $|a-b|\lt 5$, then $|b|\lt|a|+5.$
I've first written down $-5\lt a-b \lt5$ and have tried to add different things from all sides of the inequality. Like adding $b+5$ to get $b\lt a+5 \lt 10+b$ but am just not seeing where that gets me.
 A: Hint:
Use the reverse triangle inequality:
$$||x|-|y||\leq |x-y|~\forall~x,y\in\Bbb{R}$$
A: Notice that $|b|=|b+a-a|=|(a)+(b-a)|$. Now use the triangle inequality.
A: If this exercise was intended to be solved without the triangle inequality, 
If $|a - b| \lt 5$ then by isolating $b$ you will get that $ a - 5 \lt b \lt a + 5 $ and then multiplying by $-1$ you get $ -a -5 \lt -b \lt 5 - a $. 
So you can draw out $ b \lt a + 5$ and $ -b \lt 5 - a $. So combining these last two inequalities, 
$$| b| \lt \max\{ 5 - a, 5 + a \} \tag{1}$$
Now use the fact that $ a \le |a| \;\ \text{and} -a \le |a| $ to obtain the desired result. result
A: If $|a-b|<5$, then $|b|=|b-a+a|\le |b-a|+|a| < 5+|a|$. That's it!
A: Maybe this gives a different point of view.  Start by supposing that, to the contrary, $|b| > |a| + 5 $.  We then want to show that this implies $|a - b | > 5$.
Just go case by case.  
Case I: $b > 0, a > 0$.
Then the assumption implies $b > a + 5$ which implies $b - a > 5$ which implies 
$|b - a| > 5$ as desired.
Case II.  $b < 0, a < 0$.
The assumption implies $-b > -a + 5$ which implies $a - b > 5$ so again $|a - b| > 5$
Case III.  $b > 0, a < 0$
The assumption implies $b > -a + 5$.  Add $-a$ (a positive quantity) to both sides to get $b - a > -a + 5$ which implies $|b - a| > -2a + 5 > 5$ as desired.
Case IV.  $b < 0, a > 0$.
The assumption implies $-b > a + 5$.  Add $a$ (a positive quantity) to both sides to get $a - b > 2a + 5$ which implies $|a -b| > 2a + 5 > 5$ as desired.
The cases where either (or both) are $0$ are trivial.
A: Some general properties of the absolute value
First let me recall that
$$ |x| :=\begin{cases}x & \text{if }x \geq 0,\\ -x & \text{if }x <0.\end{cases}$$
We show that $|x|=|-x|$ for every $x\in \Bbb R$:


*

*If $x\geq 0$ then $|x| = x$ and $|-x|= -(-x)=x$.

*If $x< 0$ then $|x| = -x$ and $|-x|= -x$.


$$ \rightarrow \qquad \boxed{\quad |x|=|-x| \qquad \phantom\int\forall x \in \Bbb R\quad} \tag{1}$$
Now, we show that $x\leq |x|$ for every $x\in\Bbb R$:


*

*If $x\geq 0$, then $|x| = x$ and so $x\leq |x|$.

*If $x< 0$, then $|x| = -x$ and so $x <0 < -x = |x|$.


$$ \rightarrow \qquad \boxed{\quad x\leq |x| \qquad\phantom\int \forall x \in \Bbb R\quad }\tag{2}$$
As consequence of $(1)$ and $(2)$ for any $x\in\Bbb R$, we get
$$ -x \overset{(2)}{\leq} |-x| \overset{(1)}{=} |x|.$$
$$ \rightarrow \qquad \boxed{\quad-x\leq |x| \qquad\phantom\int \forall x \in \Bbb R \quad}\tag{3}$$
The last tool we need is that for every $x\in \Bbb R$ and every $t> 0$, if $|x|<t$ then $x<t$:


*

*If $x \geq 0$, then $|x|=x$ and thus $x=|x|<t$.

*If $x<0$, then $x<0< t$.


$$ \rightarrow \qquad \boxed{\quad\forall x \in \Bbb R \ \text{ and }\  \forall t> 0:\phantom\int |x|<t \implies x<t\quad }\tag{4}$$
The proof you want:
Now suppose that $a,b\in \Bbb R$ are such that $|a-b|< 5$. Then we have the following relations:
$$|a-b|< 5 \overset{(4)}{\implies} a-b<5 \overset{(*)}{\implies} -b<5-a=5+(-a) \overset{(3)}{\leq} 5+|a|  $$
and thus $-b<5+|a|$ (for step $(*)$, subtract $a$ from both sides of the inequality). Moreover,
$$|a-b|< 5 \overset{(1)}{\implies} |b-a|<5 \overset{(4)}{\implies} b-a<5 \overset{(**)}{\implies} b< 5+a \overset{(2)}{\leq}5+|a|$$
and thus $b< 5 +|a|$ (for step $(**)$, add $a$ to both sides of the inequality). Finally
$$-b<5+|a| \quad\text{ and }\quad b< 5 +|a| \qquad\implies\qquad |b|<5+|a|$$
(because either $|b|=b$ or $|b| =-b$).
Generalizaion:
$$ \boxed{\quad\forall a,b \in \Bbb R \ \text{ and }\  \forall t>0:\phantom\int |a-b|<t \implies |b|<t+|a|\quad }$$
Proof 1: Replace $5$ by $t$ in the proof above.
Proof 2: We can also use the triangle inequality as already suggested by the other answers above.
We show that $|x+y|\leq |x|+|y|$ for every $x,y\in\Bbb R$:
We have
$$ x+y \overset{(2)}{\leq} |x|+y \overset{(2)}{\leq} |x|+|y|\qquad
\text{and}\qquad-(x+y)=(-x)+(-y) \overset{(3)}{\leq} |x|+(-y) \overset{(3)}{\leq} |x|+|y|.$$
$$ \rightarrow\qquad\boxed{\quad \phantom\int |x+y|\leq |x|+|y| \qquad\forall x,y \in \Bbb R\quad }\tag{5}$$
It follows that
$$|b| =|a+(b-a)| \overset{(5)}{\leq} |a|+|b-a| \overset{(***)}{<} |a|+t,$$
where we have used that $|b-a|\overset{(1)}{=}|a-b|<t$ for step $(***)$.
A: $$||a|-|b||\leq|a-b|<5$$
by the reverse triangle inequality. 
$$|b|-|a|\leq||a|-|b||<5$$
Because $-x \leq |x|$ for all $x$.
$$|b|-|a|<5$$
$$|b|<5+|a|$$
A: We will use reverse triangle inequality i.e. $||a|-|b|| \le |a-b|$ and what you got, that $|a-b|<5 \implies -5<a-b<5$
Note : $x\le y<z \implies x<z$ because if the equality holds and $x=y$ then too $x<z$.
$$|a-b|<5
\\ \implies ||a|-|b|| \le |a-b| <5  
\\ \implies ||a|-|b|| < 5
\\ \implies -5 < |a|-|b|
\\ \implies |b| < |a|+5$$
A: The main problem is that the inequality is NOT 'implied and implied by' ($\iff$). This means that just because we have:
$$|a-b|\lt5\implies |b|\lt|a|+5$$
this doesn't mean that:
$$|b|\lt|a|+5 \implies |a-b|\lt5$$
as a global truth is true, in fact it is false, for example let $a=4, b=-3$.
We can see that the first condition is necessarily true from a diagram:

The green regions
shows the area for which $|a-b|\lt5$, and the orange regions show for which regions $|b|\ge|a|+5$ (except for they both use solid lines - you have to imagine the boundaries as open or closed!).
The white regions are therefore regions where $|b|\lt|a|+5$ but $|a-b|\ge5$.
A: We have that $$|a-b| \leq 5$$ 
It stands that $$||a|-|b|| \leq |a-b| \\ \Rightarrow -|a-b| \leq |a|-|b| \leq |a-b|$$ 
From the inequalities $-|a-b| \leq |a|-|b|$ and $|a-b| \leq 5$ we get $$|b|-|a| \leq |a-b| \leq 5 \Rightarrow |b|-|a| \leq 5 \Rightarrow |b| \leq |a|+5$$ 
