Условие задачи
Решите неравенство: \(\sqrt {log_3x } + 2\sqrt {log_x3 } \geq 3\)
Решение
ОДЗ: \(\displaystyle \left\{\begin{gathered} x\, \textgreater \, 0\\ x \ne 1 \\ log_x 3 \, \textgreater \, 0 \end{gathered}\right.\)
\(\sqrt{log_3 x} + 2\sqrt{log_x 3}\geq 3\)
Замена: \(\sqrt{log_3 x} = t, \, t \, \textgreater \, 0, \) тогда
\(\displaystyle \sqrt{log_x 3} = \frac{1}{t},\)
\(\displaystyle t+ \frac{2}{t} \geq 3,\)
\((t-1)(t-2) \geq 0,\) отсюда \(\displaystyle \left[ \begin{gathered} t \geq 2 \\ t \leq 1 \\ \end{gathered} \right.\)
Вернёмся к переменной x.
\(\displaystyle \left[ \begin{gathered} 0 \, \textless \sqrt{log_3 x} \leq 1 \\ \sqrt{log_3 x} \geq 2 \\ \end{gathered} \right.\)
\(\displaystyle \left[ \begin{gathered} 1\, \textless x \leq 3 \\ x \geq 81 \\ \end{gathered} \right.\)
Ответ
\(x \in (1;3] \cup [81; + \infty ). \)
