How to solve this inequality $$\sqrt{2+x}\ge |x|-1$$ 
How to solve such question that include both absolute value and square root? I have tried solving absolute value and then $\sqrt{2+x}\ge0$
 A: for the square root we have the condition $$x\geq -2$$
then if $$|x|\le 1$$  inequality is fulfilled, in the other case $$|x|\geq 1$$ we Can square both sides we get
$$0\le x^2-2|x|-x-1$$
for $$x\geq 0$$ we have to solve $$0\le x^2-3x-1$$ this gives 
$$x\le \frac{1}{2} \left(3+\sqrt{13}\right)$$
for $$x<0$$ we have
$$x^2+x-1\geq 0$$
this gives $$x\geq \frac{1}{2} \left(-1-\sqrt{5}\right)$$
finally we get
$$\frac{1}{2} \left(-1-\sqrt{5}\right)\leq x\leq
   \frac{1}{2} \left(3+\sqrt{13}\right)$$
A: HINT
There are not shortcuts when we deal with absolute values and square roots we need to consider different cases and then put all solution togheter.
In this case note that we have solutions only if $x+2\ge0 \implies x\ge-2$ and we need to consider 2 main cases
1) $x\ge0 \implies \sqrt{2+x}\ge x-1$


*

*for $x-1<0 \implies 0\le x<1$ inequality always holds since LHS is
positive

*for $x-1\ge 0 \implies x\ge 1$ we can square both sides and obtain
$2+x\ge (x-1)^2$


2) $-2\le x<0 \implies \sqrt{2+x}\ge -x-1$


*

*for $-x-1<0 \implies -1\le x<0$ inequality always holds since LHS is
positive

*for $-x-1\ge 0 \implies -2 \le x\le -1$ we can square both sides and obtain
$2+x\ge (-x-1)^2$

A: Consider the intervals $x<-2$, $-2<x<0$,and,$x>0.$
The inequality changes accordingly.
A: If $|x|\leqslant1$, then it is obvious that$$\sqrt{2+x}\geqslant|x|-1\tag1$$holds. Otherwise, consider two cases: $x>1$ and $x<-1$. In the first case, $(1)$ becomes $\sqrt{2+x}\geqslant x-1$, which is equivalent to $2+x\geqslant(x-1)^2$. Find the solutions such that $x>1$. Now, deal with the case $x<-1$ in a similar way.
