Let $P,Q$ be quadratic polynomials with discriminants $p,q$ such that $|P(x)|\geq|Q(x)|$ for all real numbers $x$. Prove that $|p|\geq |q|$.
Suppose that $P(x)=ax^2+bx+c$ and $Q(x)=dx^2+ex+f$. Then $p=b^2-4ac$ and $q=e^2-4df$. If $P(x)\geq Q(x)$ for all $x$, we have $P(x)-Q(x)\geq 0$, which means its discriminant $$(b-e)^2-4(a-d)(c-f)<0.$$ Expanding, this becomes $$p+q=(b^2-4ac)+(e^2-4df)<2(be-2cd-2af).$$ How can we go from here?
I think this is not true. For example, let $$P(x)=x^2+1, Q(x)=x^2-1$$ with $p=-4, q=4$. Clearly $|P(x)|\geq |Q(x)|$ for all real number $x$, however $p\ngeq q$.
To prove $|p|\geq |q|$. Assume $P(x)=ax^2+bx+c, Q(x)=dx^2+ex+f$, and $a>0$.
Case 1: If $p<0$.
Then $|P(x)|\geq|Q(x)|\Rightarrow P(x)\geq |Q(x)|\Rightarrow P(x)\geq\pm Q(x)$ for all real number $x$. $$\Longrightarrow(a\pm d)x^2+(b\pm e)x+c\pm f\geq 0$$ $$\Longrightarrow p+q\pm 2(be-2dc-2af)\leq 0$$ i.e. $|p|=-p\geq q+2(be-2dc-2af)$ and $|p|=-p\geq q-2(be-2dc-2af)$ $$\Longrightarrow|p|\geq q\Longrightarrow|p|\geq|q|$$
Case 2: If $p<0.$
Then $P(x)$ has two distinct zeros, say $x_1,x_2$, which implies $Q(x_1)$ and $Q(x_2)$ must be zeros. $$\Longrightarrow P(x)=\frac{a}{d}Q(x)\Longrightarrow|\frac{a}{d}|\geq1\Longrightarrow|p|=\frac{a}{d}|q|\geq|q|.$$
Case 3: If $p=0$.
Then we can write $P(x)=a(x-x_0)^2$, and $Q(x)=d(x-x_0)(x-x_0')$ for some zeros $x_0$ and $x_0'$ $$|P(x)|\geq|Q(x)|\Longrightarrow |a||x-x_0|\geq |d||x-x_0'|$$ for all real number $x$. Thus, $x_0'$ must be equal to $x_0$. Then we get $q=0$.
In conclude, we always have $|p|\geq|q|$.