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Good morning to everyone. I have an inequality and I have to find the sign of it. At first I thought that it's positive on all intervals (excepting the number $2$), but after looking on my answer sheet I saw that on an interval it isn't, and I don't know how to deduce this interval. The inequality is the following: $$ 2\sqrt{x^2-4}-x+1 \ge 0 \rightarrow x \ge 2$$ $$ 2\sqrt{4-x^2}-x+1\ge 0 \rightarrow 0 < x < 2 $$ For the first the function is positive for all $ x \ge \frac{-1+2\sqrt{13}}{3}$ and for $ (2,\frac{-1+2\sqrt{13}}{3}) $ (Here I just assume that it's positive); we obtain that it's negative in $x=2$.

The second function is positive for all $ x \in (0,\frac{1+2\sqrt{19}}{5})$ and I also assume that it's positive for all $ x \in (\frac{1+2\sqrt{19}}{5},2)$. (These are all my calculations)

Instead I don't know how they found that it is negative for $ \frac{1+2\sqrt{19}}{5} < x < \frac{-1+2\sqrt{19}}{5}$ It doesn't even make sense because the first number is bigger then the second.

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  • $\begingroup$ At one point or another, you will have to note that the first inequality is also $2\sqrt{x^2-4}\geqslant x-1$, always true when $x\leqslant-2$, and, for $x\geqslant2$, equivalent to $4(x^2-4)\geqslant x^2-2x+1$, and then to solve the sign of the quadratics $4(x^2-4)-(x^2-2x+1)=3x^2+2x-17$. I doubt there is anything more to this stuff... $\endgroup$ – Did Jul 4 '16 at 9:45
  • $\begingroup$ @Did yeah I already did that and I found that it's positive for $x \ge \frac{-1+2\sqrt{13}}{3}$ $\endgroup$ – T4yl0r Jul 4 '16 at 9:54

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