\(P=\dfrac{3}{4}x+\dfrac{1}{x}+\dfrac{2}{y^2}+y=\left(\dfrac{x}{4}+\dfrac{1}{x}\right)+2\left(\dfrac{1}{y^2}+\dfrac{y}{8}+\dfrac{y}{8}\right)+\dfrac{1}{2}\left(x+y\right)\)
\(P\ge2\sqrt{\dfrac{x}{4x}}+2.3\sqrt[3]{\dfrac{y^2}{64y^2}}+\dfrac{1}{2}.4=\dfrac{9}{2}\)
Dấu "=" xảy ra khi \(x=y=2\)
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