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[trang2004]

Cho x y z>0 và $\sqrt{xy}+\sqrt{yz}+\sqrt{xz}=1$Tìm Min P= $\sum \frac{x^3}{y(x+z)}$

[conankun]

Theo AM-GM

$\frac{x^3}{y(x+z)}+\frac{y}{2}+\frac{x+z}{4}\geq \frac{3x}{2}$

Thiết lập tương tự ta sẽ có: $\sum \frac{x^3}{y(x+z)}\geq \frac{x+y+z}{2}\geq \frac{\sum \sqrt{xy}}{2}=\frac{1}{2}$

Edited by conankun, 27-06-2018 - 22:52.