Đáp án đúng: B
Giải chi tiết:Đặt \(\left\{ \begin{array}{l}{\log _2}x = a\\{\log _2}y = b\\{\log _2}z = c\end{array} \right.\), khi đó ta có \(xyz = {2^a}{.2^b}{.2^c} = 1 \Leftrightarrow a + b + c = 1\).
Khi đó ta có: \(S = \log _2^3x + \log _2^3y + \dfrac{1}{4}\log _2^3z \Rightarrow S = {a^3} + {b^3} + \dfrac{1}{4}{c^3}\).
Ta có \(c = 1 - a - b\), thay vào \(S\) \( \Rightarrow S = {a^3} + {b^3} + \dfrac{1}{4}{\left( {1 - a - b} \right)^3}\)
Ta có
\(\begin{array}{l}{a^3} + {b^3} = {\left( {a + b} \right)^3} - 3ab\left( {a + b} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ge {\left( {a + b} \right)^3} - 3.\dfrac{{{{\left( {a + b} \right)}^2}}}{4}\left( {a + b} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ge {\left( {a + b} \right)^3} - \dfrac{3}{4}{\left( {a + b} \right)^3} = \dfrac{1}{4}{\left( {a + b} \right)^3}\end{array}\)
Khi đó ta có
\(\begin{array}{l}S \ge \dfrac{1}{4}{\left( {a + b} \right)^3} + \dfrac{1}{4}{\left( {1 - a - b} \right)^3}\\S \ge \dfrac{1}{4}\left[ {{{\left( {a + b} \right)}^3} + {{\left( {1 - a - b} \right)}^3}} \right]\\S \ge \dfrac{1}{4}.\dfrac{1}{4}{\left( {a + b + 1 - a - b} \right)^3}\\S \ge \dfrac{1}{{16}}\end{array}\)
Vậy \({S_{\min }} = \dfrac{1}{{16}}\).
Chọn B.
