[Toán 9]

L

lp_qt

$A=\dfrac{a^3+1}{a^2}=a+\dfrac{1}{a^2}$

$=\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{1}{a^2} \ge 3.\sqrt[3]{\dfrac{a}{2}.\dfrac{a}{2}.\dfrac{1}{a^2}}=\dfrac{3}{\sqrt[3]{4}}$

khi $\dfrac{a}{2}=\dfrac{1}{a^2} \Longleftrightarrow a=\sqrt[3]{2}$
 
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