Cho x,y là các số thực không âm thỏa mãn $x^{2}y=1$. Tìm min $x^{2}+4xy$
Tìm min $x^{2}+4xy$
Started By duyanh782014, 23-02-2018 - 11:27
#1
Posted 23-02-2018 - 11:27
#2
Posted 23-02-2018 - 16:56
$x^{2}+4xy=x^{2}+\frac{4}{x} =x^{2}+\frac{2}{x}+\frac{2}{x}\geq 3\sqrt[3]{4}<=> x=\sqrt[3]{2} ,y=\frac{1}{\sqrt[3]{4}}$
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