<s>[INST] can we have a triangle with sides $1, x$ and $x^2$? Can we have a triangle with sides $1$, $x$ and $x^2$? And what could $x$ be?
I try to approach this question by making 3 inequalities.
$1+x>x^2$,
$1+x^2>x$,
$x^2+x>1$  
and they come with different quadratic inequalities
$x^2-x-1<0$ (solution is $(1+\sqrt 5)/2 > x > (1-\sqrt 5)/2$ ) ;
$x^2-x+1>0$ (solution is $x > (-1+\sqrt 5)/2$ or $x < (-1-\sqrt 5)/2$ )
$x^2+x-1>0$ (no real solution)
Then I start to struggle with the next step.... Thank you
 [/INST] So you already have the inequalities:
$$1+x>x^2\Longleftrightarrow x^2-x-1<0\Longleftrightarrow  \frac{1-\sqrt 5}{2}<x<\frac{1+\sqrt 5}{2}$$
$$1+x^2>x\Longleftrightarrow x^2-x+1>0\Longrightarrow\,\text{always true}$$
$$x^2+x>1\Longleftrightarrow x^2+x-1>0\Longleftrightarrow x<\frac{-1-\sqrt 5}{2}\,\,\vee\,\,x>\frac{-1+\sqrt 5}{2}$$
The common interval to all the above inequalities is
$$\frac{-1+\sqrt 5}{2}<x<\frac{1+\sqrt 5}{2}$$
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