Cho x y z >0 và xyz=1. Tìm min P=$\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+\frac{3}{x+y+z}$
Cho x y z >0 và xyz=1. Tìm min P=$\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+\frac{3}{x+y+z}$
#1
Đã gửi 19-06-2018 - 21:08
#2
Đã gửi 19-06-2018 - 21:12
$xyz=1=>P=x+y+z+\frac{3}{x+y+z}=\frac{x+y+z}{3}+\frac{3}{x+y+z}+\frac{2(x+y+z)}{3}\geq 2\sqrt{\frac{x+y+z}{3}.\frac{3}{x+y+z}} +\frac{2.3\sqrt[3]{xyz}}{3}=2+2=4<=>x=y=z=1$
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#3
Đã gửi 19-06-2018 - 21:12
Cho x y z >0 và xyz=1. Tìm min P=$\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+\frac{3}{x+y+z}$
$S=x+y+z+\frac{3}{x+y+z}=\frac{x+y+z}{3}+\frac{3}{x+y+z}+\frac{2(x+y+z)}{3}\geq 2+\frac{2.3\sqrt[3]{xyz}}{3}=4$
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