2 replies to this topic

[Dam Uoc Mo]

Cho x,y,z>0 và $\sqrt{x}+\sqrt{y}+\sqrt{z}=1 . \\ CMR: \frac{x^{2}+yz}{\sqrt{2x^{2}(y+z)}}+\frac{y^{2}+zx}{\sqrt{2y^{2}(z+x)}}+\frac{z^{2}+xy}{\sqrt{2z^{2}(x+y)}}\geq 1.$

Edited by Dam Uoc Mo, 13-12-2014 - 16:26.

[vipboycodon]

Xem ở http://diendantoanho...hức-thi-dại-họ/

Edited by vipboycodon, 13-12-2014 - 16:38.

[Algebra]

Cho x,y,z>0 và $\sqrt{x}+\sqrt{y}+\sqrt{z}=1 . \\ CMR: \frac{x^{2}+yz}{\sqrt{2x^{2}(y+z)}}+\frac{y^{2}+zx}{\sqrt{2y^{2}(z+x)}}+\frac{z^{2}+xy}{\sqrt{2z^{2}(x+y)}}\geq 1.$

$\frac{x^{2}+yz}{\sqrt{2x^{2}(y+z)}}+\frac{y^{2}+zx}{\sqrt{2y^{2}(z+x)}}+\frac{z^{2}+xy}{\sqrt{2z^{2}(x+y)}}= \frac{1}{\sqrt{2}}(\sum \frac{x}{\sqrt{y+z}}+\sum \frac{yz}{x\sqrt{y+z}})$

$\geq \frac{1}{\sqrt{2}}(\frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^{2}}{\sum \sqrt{x+y}}+\frac{1}{xyz}(\sum \frac{y^{2}z^{2}}{\sqrt{y+z}}))\geq \frac{1}{\sqrt{6(x+y+z)}}+\frac{(xy+yz+zx)^{2}}{xyz\sqrt{6(x+y+z)}}\geq \frac{1}{\sqrt{6(x+y+z)}}+\frac{3xyz(x+y+z)}{xyz\sqrt{6(x+y+z)}}= \frac{1}{\sqrt{6(x+y+z)}}+\sqrt{\frac{3(x+y+z)}{2}}\geq 1$

Edited by Algebra, 16-12-2014 - 21:50.