Chủ đề này có 6 trả lời

[Phuongthaonguyen]

1) Cho a,b,c>0, abc=1. CMR: $\frac{a}{ab+1}+\frac{b}{bc+1}+\frac{c}{ca+1}\geq \frac{3}{2}$

2)Cho a,b,c>0 thỏa abc=1. CMR: $\frac{1}{a^{2}(b+c)}+\frac{1}{b^{2}(a+c)}+\frac{1}{c^{2}(a+b)}\geq \frac{3}{2}$

3)Cho 3 số dương x,y,z có tổng bằng 1. CMR: $\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\geq 1+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}$

[Kiratran]

3,https://diendantoanh...-thỏa-mãn-xyz1/

[DOTOANNANG]

Sử dụng bất đẳng thức Minkowski:

 

$$\sqrt{(\,\sqrt{\,x})^{\,2}\,+\,(\sqrt{\,yz})^{\,2}}\,+\, \sqrt{\,(\sqrt{\,y})^{\,2}\,+\,(\,\sqrt{\,zx})^{\,2}}\,+\,\sqrt{\,(\sqrt{\,z})^{\,2}\,+\,(\sqrt{\,xy})^{\,2}}\,\geq$$

 

$$\geq\,\sqrt{(\sqrt{\,x}\,+\,\sqrt{\,y}\,+\,\sqrt{\,z})^{\,2}\,+\,(\sqrt{\,xy}\,+\,\sqrt{\,yz}\,+\,\sqrt{\,zx})^{\,2}}\,=$$

 

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

 

 

[HelpMeImDying]

2) bđt $\Leftrightarrow \frac{bc}{ab+ac}+\frac{ca}{ba+bc}+\frac{ab}{ca+cb}\geq \frac{3}{2}$

đúng theo bđt Nestbit

[Kiratran]

1, đặt $\frac{x}{y}=a, \frac{y}{z}=b,\frac{z}{x}=c$

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

sd $(xy+yz+xz)^2 \geq 3xyz(x+y+z)$ => đpcm

[Phuongthaonguyen]

Sử dụng bất đẳng thức Minkowski:

 

$$\sqrt{(\,\sqrt{\,x})^{\,2}\,+\,(\sqrt{\,yz})^{\,2}}\,+\, \sqrt{\,(\sqrt{\,y})^{\,2}\,+\,(\,\sqrt{\,zx})^{\,2}}\,+\,\sqrt{\,(\sqrt{\,z})^{\,2}\,+\,(\sqrt{\,xy})^{\,2}}\,\geq$$

 

$$\geq\,\sqrt{(\sqrt{\,x}\,+\,\sqrt{\,y}\,+\,\sqrt{\,z})^{\,2}\,+\,(\sqrt{\,xy}\,+\,\sqrt{\,yz}\,+\,\sqrt{\,zx})^{\,2}}\,=$$

 

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

cái đoạn cuối cùng là sao vậy

[thanhdatqv2003]

là hằng đẳng thức do