Chứng minh rằng \({\log _2}3 > {\log _3}4\)
Ta có:
\(\begin{array}{l}
{\log _2}3 > {\log _3}4\\
\Leftrightarrow \frac{1}{{{{\log }_3}2}} > {\log _3}4\\
\Leftrightarrow {\log _2}.{\log _3}4 < 1
\end{array}\)
(vì \({\log _3}2 > 0\))
Áp dụng BĐT Cô - si cho 2 số dương ta có:
\(\begin{array}{*{20}{l}}
\begin{array}{l}
\sqrt {{{\log }_3}2.{{\log }_3}4} \\
< \frac{1}{2}({\log _3}2 + {\log _3}4) = \frac{1}{2}{\log _3}8\\
< \frac{1}{2}{\log _3}9 = 1
\end{array}\\
{ \Rightarrow {{\log }_3}2.{{\log }_3}4 < 1(dpcm)}
\end{array}\)
-- Mod Toán 12
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