2 replies to this topic

[ILoveMath4864]

cho x, y, z lớn hơn 0 thỏa mãn $\sqrt{x}+\sqrt{y}+\sqrt{z}=3$

Tìm Min của $P=\frac{x\sqrt{x}}{x+y}+\frac{y\sqrt{y}}{y+z}+\frac{z\sqrt{z}}{z+x}$

[loolo]

$(\sqrt{x},\sqrt{y},\sqrt{z})\rightarrow (a,b,c)$

Theo đề ta có: $a+b+c=3$

$P=\sum \frac{a^{3}}{a^{2}+b^{2}}=\sum (a-\frac{ab^{2}}{a^{2}+b^{2}})\geq \sum (a-\frac{ab^{2}}{2ab})=\sum a-\sum \frac{b}{2}=\sum a=3/2$

Dấu " = " tại $a=b=c=1\Leftrightarrow x=y=z=1$

Edited by loolo, 22-09-2016 - 20:39.

[basketball123]

$=\sqrt{x}-\frac{y\sqrt{x}}{x+y}+\sqrt{y}-\frac{z\sqrt{y}}{y+z}+\sqrt{z}-\frac{x\sqrt{z}}{z+x}\geq \sqrt{x}-\frac{y\sqrt{x}}{2\sqrt{xy}}+\sqrt{y}-\frac{z\sqrt{y}}{2\sqrt{yz}}+\sqrt{z}-\frac{x\sqrt{z}}{z+x}=\sqrt{x}+\sqrt{y}+\sqrt{z}-\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{2}=\frac{3}{2}$

Dấu "=" xảy ra khi x=y=z=1