ĐK: a \(\ne\) 0
\(9a\ge\dfrac{4}{a}\)
\(\Rightarrow9a^2\ge4\)
\(\Rightarrow a^2\ge\dfrac{4}{9}\)
\(\Rightarrow\left|a\right|\ge\left|\dfrac{2}{3}\right|\)
\(\Rightarrow\left[{}\begin{matrix}a\ge\dfrac{2}{3}\left(\forall a\ge0\right)\\a\le-\dfrac{2}{3}\left(\forall a< 0\right)\end{matrix}\right.\)
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Ta có \(9a\ge\dfrac{4}{a}\)
\(9a.a\ge\dfrac{4}{a}.a\)
\(9a^2\ge4\)
\(9a^2.\dfrac{1}{9}\ge4.\dfrac{1}{9}\)
\(a^2\ge\dfrac{4}{9}\)
\(a\ge\pm\sqrt{\dfrac{4}{9}}\)
\(a\ge\pm\dfrac{2}{3}\)
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