How do I show that $\left|x+4\right|I've done proof by cases for this example, when $\left|x+4\right|\geq{0}$ and $\left|x+4\right|<0$ then...
Case I: $|x+4| = x+4 \implies x+4 < x \implies 4<0 \text{ which is impossible}$
Case II: $|x+4| = -(x+4) \implies -x-4<x \implies -2<x$ ????
I'm sort of stuck on where to go...
Edit: I realized my cases were missing some key component, namely the two cases include if $x+4\geq{0}$ then $x\geq{-4}$ and if $x+4<0$ then $x<-4$.
 A: In your case II, for $|x+4| = -(x+4)$ to hold, $x$ has to be smaller than $-4$, but you already saw that $x > -2$, so?
A: The proper casework is as follows.
If $x \ge -4$, then $x+4 \ge 0$ and $|x+4| = x+4 > x$.
If $x < -4$, then $x+4 < 0$ and $|x+4| = -(x+4)$.  Then if $-(x+4) < x$, it follows that $2x > -4$ or $x > -2$.  But this is incompatible with the assumption we made earlier that $x < -4$.
This illustrates that you cannot just take different cases and work them through:  there are additional conditions, namely whether $x \ge -4$ or $x < -4$, that are required in order for the two cases to arise in the first place.
That said, it is easier to note that for all real $x$, we have $$|x+4| \ge x+4,$$ since $|x| \ge x$ for all real $x$.  Then of course since $x+4 > x$, the result follows.
A: One more proof: it is always the case that $a \le |a|$. So,
$$x < x + 4 \le |x + 4|.$$
That is, $x < |x + 4|$, making $x > |x + 4|$ (or even $x \ge |x + 4|$) impossible.
A: Either $x\ge0$ or $x<0$.  If $x\ge0$, then $|x+4|=x+4>x$.  If $x<0$, then $x<0\le|x+4|$.
A: Hint
Clearly if $x>|x+4|,$
$$x>0\implies|x+4|=+(x+4)$$ which is definitely $>x$
A: From the inequality follows that $x^2+8x+16<x^2\iff x<-2$, hence $|x+4|<-2$.
