Lời giải:Ta có :
\(\begin{array}{l}\,\,\,\,\,\,\,\ln x + \ln y \ge \ln \left( {{x^2} + y} \right) \Leftrightarrow \ln \left( {xy} \right) \ge \ln \left( {{x^2} + y} \right)\\ \Leftrightarrow xy \ge {x^2} + y\,\,\left( {x,y > 0} \right) \Leftrightarrow y\left( {x - 1} \right) \ge {x^2} > 0\end{array}\)
Do\(y > 0\,\,\left( {gt} \right) \Rightarrow x - 1 \ge 0 \Leftrightarrow x \ge 1\), khi đó ta có \(y \ge \frac{{{x^2}}}{{x - 1}}\).
Ta có:
\(\begin{array}{l}P = 2x + y \ge 2x + \frac{{{x^2}}}{{x - 1}} = \frac{{3{x^2} - 2x}}{{x - 1}} = \frac{{3{x^2} - 3x + x - 1 + 1}}{{x - 1}} = 3x + 1 + \frac{1}{{x - 1}}\\\,\,\,\,\, = 3\left( {x - 1} \right) + \frac{1}{{x - 1}} + 4\mathop \ge \limits^{Co - si} 2\sqrt {3\left( {x - 1} \right).\frac{1}{{x - 1}}} + 4 = 2\sqrt 3 + 4\end{array}\)
Vậy \(\min P = 4 + 2\sqrt 3 \).
Chọn A.
