If $ax^2 + 2bx +c$ and $px^2 + 2qx + r$ are both $\geq 0$ prove $apx^2 + bqx + cr \geq 0$ as well

Question

$a, b, c, p, q, r \in \mathbb{R}$ such that for every real $x$:

$$\begin{equation} ax^2 + 2bx + c \geq 0 \end{equation}$$

and

$$px^2 + 2qx + r \geq 0$$

Prove that

$$apx^2 + bqx + cr \geq 0$$

I started off by plugging in some values of $x$ into the two inequalities and trying to get some constraints on the coefficients. I'm not sure if it is perfectly valid to do this, because putting in certain values of $x$ might favour certain values of other coefficients and thus give more specific constraints. Is it all right to do this?

Anyway,

Let the first inequality be denoted by $f(x) \geq 0$ and the second one by $g(x) \geq 0$

$$\implies f(0) = c \geq 0$$

$$\implies g(0) = r \geq 0$$

Another thing I noticed was, that $f(x) = 0$ when $x$ is a root of $f(x)$. If $f(x)$ has two distinct roots, then the its minimum will be negative, right? So, there must be only one distinct root.

Also, taking the derivative of $f(x)$, (I would prefer a solution without calculus):

$$2ax + 2b = 0 \implies ax = -b$$

which doesn't seem to help much at all.

Answer 1

Corrected answer.

Let me consider the case $a\neq 0,p\neq 0.$ Recall that a polynomial of degree $2$ is nonnegative if the leading coefficient is positive and discriminant is nonpositive.

First is equivalent to $a>0,\ b^2-ac\leq 0.$ Second is equivalent to $p>0,\ q^2-pr\leq 0.$ We also have $c\geq 0,r\geq 0$ (values of polynomials in $0.$) Multiplying inequalities $a>0,p>0$ and $b^2\leq ac, q^2\leq pr$ we obtain $ap>0,b^2q^2\leq acpr\leq 4acpr.$ Last two inequalities imply the positivity of $apx^2+bqx+cr.$