Đáp án:
\[\mathop {\min }\limits_{\left( {2; + \infty } \right)} f\left( x \right) = \frac{{2 + 2\sqrt 3 }}{3} \Leftrightarrow x = 2 + \sqrt 3 \]
Giải thích các bước giải:
Áp dụng bất đẳng thức AM - GM với \(x > 2\) ta có:
\(\begin{array}{l}
f\left( x \right) = \frac{x}{3} + \frac{1}{{x - 2}}\\
= \left( {\frac{x}{3} - \frac{2}{3}} \right) + \frac{1}{{x - 2}} + \frac{2}{3}\\
= \left( {\frac{{x - 2}}{3} + \frac{1}{{x - 2}}} \right) + \frac{2}{3}\\
\ge 2.\sqrt {\frac{{x - 2}}{3}.\frac{1}{{x - 2}}} + \frac{2}{3}\\
= 2.\sqrt {\frac{1}{3}} + \frac{2}{3}\\
= \frac{{2 + 2\sqrt 3 }}{3}\\
\Rightarrow f\left( x \right) \ge \frac{{2 + 2\sqrt 3 }}{3}\\
f\left( x \right) = \frac{{2 + 2\sqrt 3 }}{3} \Leftrightarrow \frac{{x - 2}}{3} = \frac{1}{{x - 2}} \Leftrightarrow x = 2 + \sqrt 3 \,\,\,\,\,\left( {x > 2} \right)
\end{array}\)
Vậy \(\mathop {\min }\limits_{\left( {2; + \infty } \right)} f\left( x \right) = \frac{{2 + 2\sqrt 3 }}{3} \Leftrightarrow x = 2 + \sqrt 3 \)
