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$$\sqrt{2+x}\ge |x|-1$$ How to solve such question that include both absolute value and square root? I have tried solving absolute value and then $\sqrt{2+x}\ge0$

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    $\begingroup$ What if $x=7$? Are we expected to prove that $3\ge 6$? $\endgroup$ – G Tony Jacobs Feb 3 '18 at 14:44
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    $\begingroup$ @GTonyJacobs then we don't include it in the solution set. I think OP is looking for x's that satisify this inequation rather than proving the inequality. $\endgroup$ – the_firehawk Feb 3 '18 at 14:49
  • $\begingroup$ Of course; that makes considerably more sense! $\endgroup$ – G Tony Jacobs Feb 3 '18 at 14:59
  • $\begingroup$ Please remember that you can choose an aswer among the given is the OP is solved, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – user Mar 9 '18 at 22:29
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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 $$\frac{1}{2} \left(-1-\sqrt{5}\right)\leq x\leq \frac{1}{2} \left(3+\sqrt{13}\right)$$

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  • $\begingroup$ according to Wolfram Alpha my solution is right, check it $\endgroup$ – Dr. Sonnhard Graubner Feb 3 '18 at 15:05
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Consider the intervals $x<-2$, $-2<x<0$,and,$x>0.$ The inequality changes accordingly.

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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 $2+x\ge (-x-1)^2$
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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.

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