Cho $a, b, c$ là các số thực thỏa mãn điều kiện $a>1, b>0, c>0$ và...

+ ${{a}^{{{x}^{2}}}}.{{\left( b+4c \right)}^{2x+3}}\ge 1$ $\Leftrightarrow {{x}^{2}}+{{\log }_{a}}\left( b+4c \right). \left( 2x+3 \right)\ge {{\log }_{a}}1=0$, $\forall x\in \mathbb{R}$ $\Leftrightarrow {\Delta }'={{\log }_{a}}^{2}\left( b+4c \right)-3{{\log }_{a}}\left( b+4c \right)\le 0$ $\Leftrightarrow 0\le {{\log }_{a}}\left( b+4c \right)\le 3$ $\Rightarrow 1\le b+4c\le {{a}^{3}}$.
+Ta có: $P=\dfrac{16a}{3}+\dfrac{1}{b}+\dfrac{{{2}^{2}}}{4c}\ge \dfrac{16a}{3}+\dfrac{{{\left( 1+2 \right)}^{2}}}{b+4c}$ $\ge \dfrac{16a}{3}+\dfrac{9}{{{a}^{3}}}=\dfrac{16a}{9}+\dfrac{16a}{9}+\dfrac{16a}{9}+\dfrac{9}{{{a}^{3}}}$
$\ge 4\sqrt[4]{{{\left( \dfrac{16}{9} \right)}^{3}}.9}=\dfrac{32}{3}$.
Đẳng thức xảy ra khi $\dfrac{16a}{9}=\dfrac{9}{{{a}^{3}}}\Rightarrow a=\dfrac{3}{2}$ và $\left\{ \begin{aligned}
& \dfrac{1}{b}=\dfrac{2}{4c} \\
& b+4c=\dfrac{27}{8} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& b=\dfrac{9}{8} \\
& c=\dfrac{9}{16} \\
\end{aligned} \right.$.
Vậy $m+n+p=\dfrac{51}{16}.$