Proving $apx^2+bqx+cr \geq 0$ for all real x

Question

If $a,b,c,p,q,r$ are real numbers such that $ax^2+bx+c \geq 0$ and $px^2+qx+r\geq 0$ for all real numbers for all real numbers the question is to prove that $apx^2+bqx+cr \geq 0$ for all real x

From the given equations I got that $a,p \gt 0$ and their discriminant is less than or equal to zero.I couldn't get how to I get the desired result using this.Any ideas?Thanks.

Answer 1

It's wrong!

Try $x^2+2x+1\geq0$ and $x^2+4x+4\geq0.$

We need $x^2+8x+4\geq0$ for all real $x$, which is wrong.