Challenging Inequalities

Consider $\displaystyle \sum_{cyc} \frac 1{a^3(b+c)} = \sum_{cyc} \frac {\frac 1{a^2}}{ab+ca} = \sum_{cyc} \frac {\frac 1{a^2}}{\frac {abc}c+\frac {abc}b} = \frac 12 \sum_{cyc} \frac {\frac 1{a^2}}{\frac 1b+\frac 1c}$ . Using Titu's lemma and then AM-GM inequality :

$\begin{aligned} \frac 12 \left( \frac {\frac 1{a^2}}{\frac 1b+\frac 1c} + \frac {\frac 1{b^2}}{\frac 1c+\frac 1a} + \frac {\frac 1{c^2}}{\frac 1a+\frac 1b} \right) & \ge \frac 12 \left(\frac {\left(\frac 1a+\frac 1b+\frac 1c\right)^2}{2\left(\frac 1a+\frac 1b+\frac 1c\right)} \right) = \frac 14 \left(\frac 1a+\frac 1b+\frac 1c\right) \ge \frac 34 \sqrt[3]{\frac 1{abc}} = \sqrt[3]{\frac {27}{128}} \end{aligned}$

Equality occurs when $a=b=c=\sqrt[3]2$ . Therefore, $2p = 2\times 27 = \boxed{54}$ .

$let~a=b=c=\sqrt[3]{2}.\\ Than~f(a,b,c)_{min}=\sqrt[3]{\dfrac{27}{128}}=\sqrt[3]{\dfrac p q}.\\ 2p=2*27=54.\\ Checked ~for~ values~ near~~ \sqrt[3]2~~ and~~ other~ values~ to~ see~ that~ it~ is~ the~ minimum.$