Đáp án đúng:
Giải chi tiết:Ta có: \(VT = \frac{{2 + 6a + 3b + 6\sqrt {2bc} }}{{2a + b + 2\sqrt {2bc} }} = \frac{{2 + 3\left( {2a + b + 2\sqrt {2bc} } \right)}}{{2a + b + 2\sqrt {2bc} }} = \frac{2}{{2a + b + 2\sqrt {2bc} }} + 3\)
Mà \(2\sqrt {2bc}  = 2\sqrt {b.2c}  \le b + 2c\) (BĐT Cô – si)
\(\begin{array}{l} \Rightarrow 2a + b + 2\sqrt {2bc}  \le 2a + b + b + 2c = 2\left( {a + b + c} \right)\\ \Rightarrow \frac{2}{{2a + b + 2\sqrt {2bc} }} \ge \frac{2}{{2\left( {a + b + c} \right)}} = \frac{1}{{a + b + c}}\\ \Rightarrow \frac{2}{{2a + b + 2\sqrt {2bc} }} + 3 \ge \frac{1}{{a + b + c}} + 3 = \frac{{{1^2}}}{{a + b + c}} + \frac{{{3^2}}}{3} \ge \frac{{{{\left( {1 + 3} \right)}^2}}}{{a + b + c + 3}} = \frac{{16}}{{a + b + c + 3}}\end{array}\)
\( \Rightarrow VT \ge \frac{{16}}{{a + b + c + 3}}\).
Ta chứng minh \(VP \le \frac{{16}}{{a + b + c + 3}}\).
Thật vậy,
\(\begin{array}{l}\frac{{16}}{{\sqrt {2{b^2} + 2{{\left( {a + c} \right)}^2}}  + 3}} \le \frac{{16}}{{a + b + c + 3}} \Leftrightarrow \sqrt {2{b^2} + 2{{\left( {a + c} \right)}^2}}  + 3 \ge a + b + c + 3\\ \Leftrightarrow \sqrt {2{b^2} + 2{{\left( {a + c} \right)}^2}}  \ge a + b + c \Leftrightarrow 2{b^2} + 2{\left( {a + c} \right)^2} \ge {\left( {a + b + c} \right)^2}\\ \Leftrightarrow 2{a^2} + 2{b^2} + 2{c^2} + 4ac \ge {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca\\ \Leftrightarrow {a^2} + {b^2} + {c^2} - 2ab - 2bc + 2ca \ge 0\end{array}\)
\( \Leftrightarrow {\left( {b - a - c} \right)^2} \ge 0\) (luôn đúng).
Do đó \(VP \le \frac{{16}}{{a + b + c + 3}}\), suy ra điều phải chứng minh.
Dấu “=” xảy ra khi \(\left\{ \begin{array}{l}b = 2c\\\frac{1}{{a + b + c}} = \frac{3}{3} = 1\\{\left( {b - a - c} \right)^2} = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = 2c\\a + b + c = 1\\b - a - c = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = \frac{1}{4}\\b = \frac{1}{2}\\c = \frac{1}{4}\end{array} \right.\).