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I'm trying to figure out this inequality:

$|x+1| + |x| \leq x^2$

I thought about trying it with two cases: $ (x = -x)$ and $(x = +x)$
but I don't seem to find out how to go through from here,

I'll be really grateful for some guidelines on what should I do,
or even so - how to approach such problem,

Thanks.

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  • $\begingroup$ Did you try squaring them? $\endgroup$ – user10444 Mar 14 '13 at 17:08
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Combine those two cases ($x > 0$ and $x < 0$) with the cases $x+1 > 0$ and $x+1 < 0$. This gives you four cases but one of them is impossible (which one?). Find an inequality for $x$ in each case, then put them together and see what you get.

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Hint: $|x+1| + |x| = |2x+1|, x> 0|x \leq -1$

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Naive approach hint:

$$(1)\;\;\;\;x<-1\;\Longrightarrow |x+1|=-x-1\;,\;\;|x|=-x\Longrightarrow -2x-1<x^2\iff $$

$$\iff x^2+2x+1>0\iff (x+1)^2>0\iff\ldots $$

$$(2)\;\;\;\;-1\le x<0\Longrightarrow |x+1|=x+1\;,\;\;|x|=-2\Longrightarrow x+1-x<x^2\iff$$

$$\iff x^2-1>0\iff (x-1)(x+1)>0\iff\ldots$$

$$(3)\;\;\;\;x\ge 0\Longrightarrow |x+1|=x+1\;,\;\;|x|=x\Longrightarrow x+1+x<x^2\iff$$

$$\iff x^2-2x-1>0\iff (x-(1+\sqrt 2))(x-(1-\sqrt 2))>0\iff\ldots$$

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