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How can I resolve this problem : $|x|-|x+1|\leq 7$ ? The answer is true for all $\mathbb{R}$ but I am having some trouble proving it.

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    $\begingroup$ Colbi's answer fleshes everything out fairly well--the only marginally "sticky" thing to realize is that $x\geq-4$ holds because you are only considering this inequality when $x\in[-1,0)$, and $x\geq-4$ obviously holds there because $-4<-1\leq x<0$. Hence, when you break the inequality up into its three corresponding pieces, it is easy to see that, together, the inequality is true for all $x\in\mathbb{R}$. $\endgroup$ – Daniel W. Farlow Jun 16 '16 at 21:34
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If $x\ge 0$ then the left side is just $1$.

If $0>x\ge -1$ then...

If $x<-1$ then...

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By the reverse triangle inequality you have that

$$\big||x|-|x+1|\big| \leq |x-(x+1)| = 1$$ and so $-1\leq|x|-|x+1|\leq 1$ for all $x\in\mathbb{R}$.

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I'll elaborate on Zachary's answer. If $x\geq 0$: $$x-(x+1)\leq 7$$ which is true $\forall ~x\in \mathbb{R} \because -1+1\leq 7+1\Rightarrow 0\leq8$.

If $x<-1$: $$-x-(-(x+1))\leq 7$$ which is true $\forall ~x\in \mathbb{R}\because -x+x+1=1\Rightarrow1\leq7$.

If $-1\leq x<0$: $$-x-(x+1)\leq7\Rightarrow 2x+1\geq-7\Rightarrow x\geq-4$$

We now have the ranges:

($x\geq0$ being true $\forall ~x\in \mathbb{R})$,($x<-1$ being true $\forall ~x\in \mathbb{R}$), ($-1\leq x<0$ and $x\geq-4$)

$\therefore |x|-|x+1|\leq 7$ is true $\forall ~x\in \mathbb{R}$.

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You can use the rule

$$|x|=\begin{cases}0\le x\to x\\x\le0\to-x.\end{cases}$$

Then

$$|x|-|x+1|=\begin{cases}0\le x&\to x-(x+1)=-1,\\-1\le x\le0&\to-x-(x+1)=-2x-1,\\x\le-1&\to-x-(-x-1)=1.\end{cases}$$

The claim holds in the middle case because the maximum value of the linear function is $-2(-1)-1<7$.

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