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Get a common denominator to write $$ \left|\frac{1}{x+1}+1\right|=\left|\frac{x+2}{x+1}\right| $$ Now it is only a matter of bounding $\frac{1}{|x+1|}$ This is a strictly increacing function on $(-\infty, -1)$. Take any $x_0\in (-2, -1)$. Then $$ \frac{1}{|x+1|} < \frac{1}{|x_0+1|} $$ for $x\in (-\infty, x_0)$. Observe that $x< x_0$ if and only if $x+2< x_0+2$. Further, $x_0+2> -2+2=0$. Set $K=x_0+2$ and $$ p=\frac{1}{|x_0+1|}. $$
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Sun, 09 Feb 2025 20:56 GMT