Đáp án: $P\ge 2$
Giải thích các bước giải:
Ta có:
$\dfrac{{11a + 11b + 2c}}{{\sqrt {8{{\rm{a}}^2} + 56} + \sqrt {8{b^2} + 56} + \sqrt {4{c^2} + 7} }} = \dfrac{{11{\rm{a}} + 11b + 2c}}{{\sqrt {8{{\rm{a}}^2} + 8\left( {ab + 2bc + 2ca} \right)} + \sqrt {8{b^2} + 8\left( {ab + 2bc + 2ca} \right)} + \sqrt {{4c^2} + ab + 2bc + 2ca} }}\\ = \dfrac{{11{\rm{a}} + 11b + 2c}}{{\sqrt {\left( {4{\rm{a}} + 4b} \right)\left( {2{\rm{a}} + 4c} \right)} + \sqrt {\left( {4{\rm{a}} + 4b} \right)\left( {2{\rm{b}} + 4c} \right)} + \sqrt {\left( {a + 2c} \right)\left( {b + 2c} \right)} }}\\ \ge \dfrac{{11{\rm{a}} + 11b + 2c}}{{\dfrac{{4{\rm{a}} + 4b + 2{\rm{a}} + 4c}}{2} + \dfrac{{4{\rm{a}} + 4b + 2{\rm{b}} + 4c}}{2} + \dfrac{{a + 2c + b + 2c}}{2}}} = 2$
Dấu = xảy ra khi
$\begin{cases}4a+4b=2a+4c\\4a+4b=2b+4c\\a+2c=b+2c\\ab+2bc+2ca=7\end{cases}$
$\to\begin{cases}a+2b=2c\\2a+b=2c\\a=b\\ab+2bc+2ca=7\end{cases}$
$\to\begin{cases}c=\dfrac32a\\a=b\\ab+2bc+2ca=7\end{cases}$
$\to\begin{cases}c=\dfrac32a\\a=b\\a^2+2a\cdot\dfrac32a+2\cdot\dfrac32a\cdot a=7\end{cases}$
$\to\begin{cases}c=\dfrac32a\\a=b\\7a^2=7\end{cases}$
$\to\begin{cases}c=\dfrac32\\a=b=1\\a=1\end{cases}$
