Given that $x^2 + ax + b > 0$ and $x^2 + (a + np)x + (b + nq) > 0$, prove that $x^2 + (a + mp)x + (b + mq) > 0, m = \overline{1, n - 1}$.

Question

Given that $$\large \left\{ \begin{align} x^2 + ax + b > 0\\ x^2 + (a + np)x + (b + nq) > 0 \end{align} \right., \forall x \in \mathbb R \ (p, q \in \mathbb R \setminus \{0\}, n \in \mathbb N, n > 1)$$

Prove that $x^2 + (a + mp)x + (b + mq) > 0, m = \overline{1, n - 1}$.

We have that $$\left\{ \begin{align} x^2 + ax + b > 0\\ x^2 + (a + np)x + (b + nq) > 0 \end{align} \right., \forall x \in \mathbb R$$

$$ \iff \left\{ \begin{align} a^2 - 4b < 0 \ (1)\\ (a + np)^2 - 4(b + nq) < 0 \ (2) \end{align} \right.$$

For $(2)$, we obtain that $p^2n^2 + 2(ap - 2q)n + (a^2 - 4b) < 0$

$$\implies (ap - 2q)^2 - p^2(a^2 - 4b) > 0 \iff q^2 - apq + bp^2 > 0 \implies \left\{ \begin{align} (ap)^2 - 4bp^2 < 0\\ (aq)^2 - 4q^2 < 0 \end{align} \right.$$

$\implies \left\{ \begin{align} a^2 - 4b < 0\\ a^2 - 4 < 0 \end{align} \right.$. And that's a dead-end.

Answer 1

Multiply the first inequality by $(n-m)$ and the second by $m$ and add these together \begin{eqnarray*} (n-m) (x^2 +ax+b) >0 \\ m(x^2+(a+np)x+(b+np))>0 \\ n(x^2+(a+mp)x+(b+mp))>0. \end{eqnarray*} Now divide the final inequality by $n$.

Answer 2

U could always check out D.See if leading coefficient >0 and D is less than 0 the quadratic is >0 for all real x.

Given : $ a^2 - 4b <0$

Given : $ (a+np)^2 - 4(b+nq)= a^2-4b + n(np^2 +2(ap-2q))<0$

(as n decreases $np^2 +2(ap-2q)$ also decreases) $\Rightarrow$ the expression is lesser for lesser 'n' and hence less than 0.