If $p, q, r$ are positive and in $AP,$ the roots of quadratic equation $p{x}^2 + px + r$ are all real for 

Given $p, q, r \in AP.$

     $2q = p + r$

Since roots are real, so

$\begin{array}{l}D\ge 0\\ \Rightarrow q^2-4pr\ge 0\\ \Rightarrow {\left(\frac{p+r}{2}\right)}^2-4pr\ge 0\\ \Rightarrow (p+r{)}^2-16pr\ge 0\\ \Rightarrow p^2+r^2-14pr\ge 0.\\ \begin{array}{r}\Rightarrow {\left(\frac{p}{r}\right)}^2-14\left(\frac{p}{r}\right)+1\ge 0\\ \Rightarrow {\left(\frac{p}{r}-7\right)}^2\ge 49-1=48\end{array}\end{array}$

$\begin{array}{r}\Rightarrow \left|\left(\frac{p}{r}-7\right)\right|\ge \sqrt{48}\\ \Rightarrow \left|\left(\frac{p}{r}-7\right)\right|\ge 4\sqrt{3}\end{array}$