$$\:{Let}\:{p}\:\&\:{q}\:{real}\:{positive}\:{number} \\$$$$\:\:{what}\:{the}\:{minimum}\:{of}\:\left(\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }\:+\frac{{q}}{{p}}\right)^{\mathrm{3}} . \\$$
$$\mathrm{let}\:{q}={pt}\wedge{t}>\mathrm{0} \\$$$$\left(\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }+\frac{{q}}{{p}}\right)^{\mathrm{3}} =\frac{\left({t}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{3}} }{{t}^{\mathrm{6}} } \\$$$$\frac{{d}}{{dt}}\left[\frac{\left({t}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{3}} }{{t}^{\mathrm{6}} }\right]=\mathrm{0}\wedge{t}>\mathrm{0}\:\Rightarrow\:{t}=\sqrt[{\mathrm{3}}]{\mathrm{2}}\:\Rightarrow\:\mathrm{min}\:\mathrm{is}\:\frac{\mathrm{27}}{\mathrm{4}} \\$$
$$\left(\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }+\frac{{q}}{\mathrm{2}{p}}+\frac{{q}}{\mathrm{2}{p}}\right)^{\mathrm{3}} \overset{{am}−{gm}} {\geqslant}\:\left(\mathrm{3}\sqrt[{\mathrm{3}}]{\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }\centerdot\frac{{q}}{\mathrm{2}{p}}\centerdot\frac{{q}}{\mathrm{2}{p}}}\right)^{\mathrm{3}} =\frac{\mathrm{27}}{\mathrm{4}} \\$$$$,,=''\:{for}\:\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }=\frac{{q}}{\mathrm{2}{p}}\Leftrightarrow{p}={q}\sqrt[{\mathrm{3}}]{\mathrm{2}}\Rightarrow{min}=\frac{\mathrm{27}}{\mathrm{4}} \\$$