A few days ago, I encountered a problem in one of my Math Textbooks. Unfortunately, there was no solution of the problem, as it was "left out" for the students to carry out on their own. The problem goes as following:
Let $$f(x)=x^2-2ax+b, \hspace{.2cm} a, b \in \mathbb{R}$$ be a quadratic function, with roots $r_1, r_2 \in \mathbb{R}$. If $|r_2|\leq |r_1|$ holds, prove that $$|a|+\sqrt{a^2+|b|}\leq(1+\sqrt{2})|r_1|$$
Now, it is trivial to prove that if $r_1=0,$ then the inequality holds, but I can't get any further than that. I have used Vieta's to express the sum and product of the roots and try to take their absolute values but to not avail. Furthermore, the expression inside the square root seems like the Discriminant but it is not quite that...
Vieta gives $$ |a| = \left| \frac{r_1+r_2}{2}\right| \le |r_1| $$ and $$ |b| = |r_1 r_2| \le |r_1|^2 $$ so that $$ |a|+\sqrt{a^2+|b|}\leq |r_1| + \sqrt{|r_1|^2 + |r_1|^2} =(1+\sqrt{2})|r_1| \, . $$