If the equation ${\mathrm{x}}^4+{\mathrm{px}}^3+{\mathrm{qx}}^2+\mathrm{rx}+5=0$ has four positive real roots, find the minimum value of $\mathrm{pr}$.

$$\begin{gathered} \mathrm{pr}=(-\mathrm{p})(-\mathrm{r}) \\ =(\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }+\mathrm{\beta })(\mathrm{\alpha \beta \gamma }+\mathrm{\alpha \beta \delta }+\mathrm{\gamma \beta \delta }+\mathrm{\gamma \delta \alpha }+\mathrm{\gamma \delta \beta }) \\ ={\mathrm{\alpha }}^2\mathrm{\beta \gamma }+{\mathrm{\alpha }}^2\mathrm{\beta \delta }+{\mathrm{\alpha }}^2\mathrm{\beta \delta }+{\mathrm{\alpha }}^2\mathrm{\gamma \delta }+\mathrm{\alpha \beta \gamma \delta }+{\mathrm{\beta }}^2\mathrm{\gamma \alpha }+{\mathrm{\beta }}^2\mathrm{\alpha \delta }+\mathrm{\alpha \beta \gamma \delta }+{\mathrm{\beta }}^2\mathrm{\gamma \delta }+{\mathrm{\gamma }}^2\mathrm{\alpha \beta }+\mathrm{\alpha \beta \gamma \delta } \\ +{\mathrm{\gamma }}^2\mathrm{\delta \alpha }+{\mathrm{\gamma }}^2\mathrm{\delta \beta }+\mathrm{\alpha \beta \gamma \delta }+{\mathrm{\alpha \beta \delta }}^2+{\mathrm{\gamma \alpha \delta }}^2+{\mathrm{\gamma \beta \delta }}^2 \\ \because \mathrm{AM}\ge \mathrm{GM} \\ \Rightarrow \frac{\mathrm{pr}}{16}\ge {\left({\mathrm{\alpha }}^{16}{\mathrm{\beta }}^{16}{\mathrm{\gamma }}^{16}{\mathrm{\delta }}^{16}\right)}^{1/6}=\mathrm{\alpha \beta \gamma \delta }=5 \Rightarrow \frac{\mathrm{pr}}{16}\ge 5 \\ \mathrm{or} \mathrm{pr}\ge 80 \\ \therefore \mathrm{Minimum} \mathrm{value} \mathrm{of} \mathrm{pr} \mathrm{is} 80. \end{gathered}$$