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Progress:

If the sides are $a$, $b$ and $c$; then $$(a,b,c)=(\frac{b}{r} ,b,br)$$ Also $$|a-c|\lt b \lt a+c \Rightarrow |\frac{b}{r} -br| \lt b \lt \frac{b}{r}+br$$

On the right side: $$b \lt \frac{b}{r}+br \Rightarrow b(\frac{1}{r}+r) \gt b \Rightarrow \frac{1}{r} +r \gt 1$$

Then I would assume $r\gt 0$ $$1+r^2 \gt r \Rightarrow r^2-r+1>0$$ This is true for every real r, so it doesn't give me any interval.

By the left side of the first inequality: $$b\gt |\frac{b}{r} -br| \Rightarrow b^2 \gt (\frac{b}{r} -br)^2 \Rightarrow b^2 \gt \frac{b^2}{r^2} -2b^2 +b^2r^2$$ Dividing everything by b, and knowing that $b \gt 0$: $$1\gt \frac{1}{r^2} -2 +r^2 \Rightarrow 1\gt (\frac{1}{r} - r)^2 \Rightarrow -1\lt \frac{1}{r} - r\lt 1$$ By the right side: $$\frac{1}{r} -r \lt 1 \Rightarrow -r^2-r+1\lt 0$$ With again no real roots, and, by the left side, we get $$-r^2+r+1\gt 0$$ again, with no real rots, and no interval.

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    Hint: let the sides be $1,r,r^2$ and write the conditions $1+r \ge r^2\,$, $r+r^2 \ge 1\,$. – 2017-02-19
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    Without loss of generality, assume $a \le b \le c$. Then $r \ge 1$. Now simply apply the triangle inequality. – 2017-02-19
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    An equilateral triangle satisfies the conditions, so that already shows that such triangles exist. – 2017-02-19
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    @dxiv: Should be strict inequality. – 2017-02-19
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    @quasi It depends on whether degenerate triangles are allowed or not. The so-called *triangle inequality* is usually written with inclusive inequalities. – 2017-02-19
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    $-r^2+r+1 = 0$ actually has real roots. – 2017-02-19
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    @dxiv: At early levels (e.g., precalculus and below), I've always seen the inequality specified as a strict inequality. – 2017-02-19
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    @quasi That's a fair point. – 2017-02-19
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    Duplicate of (http://www.cut-the-knot.org/m/Geometry/SidesInGeometricProgression.shtml) – 2017-02-20

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let $$a,b=aq,c=aq^2$$ the three sides of a triangle in a geometric progression, the triangle inequality states that $$|a-b|0$$ we get $$|1-q|