Let $a x^2 + b x +c =0$ be a quadratic equation where $|a|> |a+b+c| $, $a,b,c \in \mathbb{R}$. Prove that this equation has at least one solution $ z \in \mathbb{C}$ such that $|z| < 2$.
I don't know how to prove this, I noticed that $|a+b+c | = |f(1)|$, so $|a|> |f(1)|$, but I'm not sure if it helps. Also, $|z|= \frac{b^2}{4a^2} + \frac{ 4ac - b^2}{4 a^2} <4\implies a c <4 a^2 $. Can anyone help? I would be greatly thankful.
If both roots $s,t$ are real then we have $a(x-s)(x-t)=ax^2-a(s+t)x+ast=0$. Now $$|a|>|a-a(s+t)+ast|\implies1>|1-s-t+st|=|(1-s)(1-t)|.$$ Clearly $|s|$ and $|t|$ cannot both be greater than $2$. The double root case is straightforward.
If both roots $s=u+iv$ and $t=u-iv$ are complex we have $$a(x-(u+iv))(x-(u-iv))=ax^2-2aux+a(u^2+v^2)=0.$$ Now $$|a|>|a-2au+a(u^2+v^2)|\implies1>|1-2u+(u^2+v^2)|.$$ If $|s|$ (and hence $|t|$) are greater than $2$ then $u^2+v^2>4$. Can you proceed?