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Let $f$ and $g$ be a quadratic funtcions. Assume that $|f(x)|\geq |g(x)|$ for all $x\in\mathbb{R}$. How to show that $|d_f|\geq|d_g|$, where $d_h$ denote the discriminant of an arbitrary quadratic function $h$?

guilherme
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  • Let $f(x)=a_fx^2+b_fx+c_f$ and $g(x) = a_gx^2+b_gx+c_g$.
    Every root of $f$ has to be a root of $g$, since for every root $x_0$ $$0=|f(x_0)|\geq|g(x0)|\geq 0 \implies g(x_0)=0$$ Also, if $d_f$ is positive, so must be $d_g$. And $|a_f|/|a_g|= \lim_{x\to \infty}{|f(x)|/|g(x)|} \geq 1$. You might have to use this and some case analysis.     – Antitheos Sep 12 '15 at 22:20

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