$\left.\begin{matrix} a^2+\left(\dfrac{19-\sqrt{37}}{12} \right)^2 \ge 2\left(\dfrac{19-\sqrt{37}}{12} \right)a\\ b^2+\left(\dfrac{19-\sqrt{37}}{12} \right)^2 \ge 2\left(\dfrac{19-\sqrt{37}}{12} \right)b\\c^3+\left( \dfrac{-1+\sqrt{37}}{6}\right)^3+\left( \dfrac{-1+\sqrt{37}}{6}\right)^3 \ge3 \left( \dfrac{-1+\sqrt{37}}{6}\right)^2c \end{matrix}\right\}$Lại có $2\left(\dfrac{19-\sqrt{37}}{12} \right)=3 \left( \dfrac{-1+\sqrt{37}}{6}\right)^2$
$\Rightarrow P\ge 2\left(\dfrac{19-\sqrt{37}}{12} \right)^2+\left( \dfrac{-1+\sqrt{37}}{6}\right)^3 $