Given that $\left|a\left(b-c\right)\right|>\left|b^2-ca\right|+\left|c^2-ab\right|$ and the equation $ax^2+bx+c=0$ has at least a real root. Prove that at least one real root of $ax^2+bx+c=0$ is in $(0;\sqrt{3}-1)$.
I tried to solve the problem without success:
Letting $b'=\frac{b}{a}, c'=\frac{c}{a}$, I'm left with $$x^2+b'x+c'=0(1)$$ and $$\left|\left(b'-c'\right)\right|>\left|b'^2-c'\right|+\left|c'^2-b'\right|(2).$$
Since $f(x)=x^2+b'x+c'$ is continuous on $\mathbb{R}$ I decided to prove that $f(0)f(\sqrt{3}-1)<0$, from which $x_0 \in (0;\sqrt{3}-1)$ follows.
From (1) I have $b'^2-4c'\geq 0$. And from (2) I have that $$\left|\left(b'-c'\right)\right|>\left|b'^2+c'^2-b'-c'\right|>\left|b'^2+c'^2\right|-\left|b'+c'\right|,$$ which means $$|b'-c'|+|b'+c'|>|b'^2+c'^2|\Leftrightarrow 2(b'^2+c'^2)>(b'^2+c'^2)^2\Leftrightarrow b'^2+c'^2<2.$$
From this I can deduce a few things like $|b'c'|<1, b'+c'<2$, or even $-2-\sqrt{6}<c'<-2+\sqrt{6}$ but I still can't use them to prove $f(0)f(\sqrt{3}-1)<0$.
Could you help me with this problem?
Now I'll re-label the variables for convinience, $t:=x,\,x:=\frac{b}{a},\,y:=\frac{c}{a}$ ($a\ne 0$ see 1. below), so we have $$|x-y|>|x^2-y|+|y^2-x|,\quad x^2-4y\ge 0.$$ Now I'll open the absolute values signs, yes, considering $8$ possibilities (see 2. below), to solve the inequality for $x,\,y$ in order to be able to get the picture
along with the solutions $$\left[ \begin{array}{l} \begin{cases}-1<x<0\\ 1 - \sqrt{1 - x^2}<y<-x\end{cases}\\ \begin{cases}0<x<1\\ -\sqrt{-(x - 2) x}<y<-x\end{cases}\\ \end{array} \right.$$ Now the first brace (the upper region) is becoming invalid, because it's given that the equation $t^2+xt+y=0$ has real roots, thus the determinant $x^2-4y\ge 0$ i.e. $y\le \frac{x^2}{4}$, but for $-1<x<0$ $$1 - \sqrt{1 - x^2}>\frac{x^2}{4}$$ $$\sqrt{1 - x^2}<1-\frac{x^2}{4}$$ $$4\sqrt{1 - x^2}<4-x^2$$ $$16(1 - x^2)<16-8x^2+x^4$$ $$8x^2+x^4>0$$ Now I'll show that the graph of $y=-\sqrt{-(x - 2) x}$ (for $0<x<1$) is above the line $(\sqrt{3}-1)^2+(\sqrt{3}-1)x+y=0$ that completes the proof as all the valid inequality solutions will be inside the region of $\left((\sqrt{3}-1)^2+(\sqrt{3}-1)x+y\right)y<0$ $\Leftrightarrow$ $f(\sqrt{3}-1)\cdot f(0)<0$ for the $f$ defined in the OP.
$$-\sqrt{-(x - 2) x}>-\left((\sqrt{3}-1)^2+(\sqrt{3}-1)x\right)$$ $$\sqrt{-(x - 2) x}<(\sqrt{3}-1)\left((\sqrt{3}-1)+x\right)$$ $$-(x - 2) x<(\sqrt{3}-1)^2\left((\sqrt{3}-1)+x\right)^2$$ $$2 x - x^2<-2 \sqrt{3} x^2 + 4 x^2 + 12 \sqrt{3} x - 20 x - 16 \sqrt{3} + 28$$ $$(5-2 \sqrt{3})x^2+(12 \sqrt{3}-22)x- 16 \sqrt{3} + 28>0$$ $$\frac{D}{4}=(6 \sqrt{3}-11)^2-(5-2 \sqrt{3})(-16 \sqrt{3} + 28)=$$ $$4 \sqrt{3} - 7<0,$$ as $(5-2 \sqrt{3})>0$ so all the parabola is above $y=0$ (because no intersections), QED.
$$0>|b^2|+|c^2|$$ $$\emptyset$$
$$ \hbox{1) }\begin{cases} x−y\ge 0\\ x^2−y\ge 0\\ y^2−x\ge 0\\ \end{cases}\quad \hbox{2) }\begin{cases} x−y< 0\\ x^2−y\ge 0\\ y^2−x\ge 0\\ \end{cases}\quad \hbox{3) }\begin{cases} x−y\ge 0\\ x^2−y< 0\\ y^2−x\ge 0\\ \end{cases}\quad \hbox{4) }\begin{cases} x−y< 0\\ x^2−y< 0\\ y^2−x\ge 0\\ \end{cases}\\ \hbox{5) }\begin{cases} x−y\ge 0\\ x^2−y\ge 0\\ y^2−x< 0\\ \end{cases}\quad \hbox{6) }\begin{cases} x−y< 0\\ x^2−y\ge 0\\ y^2−x< 0\\ \end{cases}\quad \hbox{7) }\begin{cases} x−y\ge 0\\ x^2−y< 0\\ y^2−x< 0\\ \end{cases}\quad \hbox{8) }\begin{cases} x−y< 0\\ x^2−y< 0\\ y^2−x< 0\\ \end{cases}$$
$$\hbox{1) }(x-y)>(x^2-y)+(y^2-x)$$ $$y^2 + x^2-2 x+1<1$$ $$y^2 + (x-1)^2<1$$ $$\hbox{2) }-(x−y)>(x^2−y)+(y^2−x)$$ $$x^2 + y^2-2y+1<1$$ $$x^2 + (y-1)^2<1$$ $$\hbox{4) }-(x−y)>-(x^2−y)+(y^2−x)$$ $$(y - x) (x + y)<0$$ $$\hbox{5) }(x−y)>(x^2−y)-(y^2−x)$$ $$(y - x) (x + y)>0$$ $$\hbox{7) }(x−y)>-(x^2−y)-(y^2−x)$$ $$x^2 + y^2-2y+1>1$$ $$x^2 + (y-1)^2>1$$ $$\emptyset$$ $$\hbox{8) }-(x−y)>-(x^2−y)-(y^2−x)$$ $$x^2-2x+1 + y^2>1$$ $$(x-1)^2 + y^2>1$$ $$\emptyset$$