Chủ đề này có 39 trả lời

[hoangkimca2k2]

$a,b,c>0$

$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\sqrt{\frac{ab+bc+ac}{a^{2}+b^{2}+c^{2}}}\geq \frac{5}{2}$$

[DOTOANNANG]

$$\frac{x}{y+ z}+ \frac{y}{z+ x}+ \frac{z}{x+ y}+ \frac{4}{3}\,\sqrt{\frac{xy+ yz+ zx}{x^{2}+ y^{2}+ z^{2}}}\geqq \frac{17}{6}$$

[DOTOANNANG]

$$(\frac{x^t}{y^t+z^t}+\frac{y^t}{z^t+x^t}+\frac{z^t}{x^t+y^t})\,(\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}) \geqq \frac{9}{4}$$

 

$t_{\min}= \frac{1}{2}\,?$

Bài viết đã được chỉnh sửa nội dung bởi DOTOANNANG: 26-04-2018 - 10:23

[DOTOANNANG]

$$(\frac{x}{y+z}+\frac{y}{z+w}+\frac{z}{w+x}+\frac{w}{x+y})\,(\frac{x^2}{x^2+y^2}+\frac{y^2}{y^2+z^2}+\frac{z^2}{z^2+w^2}+\frac{w^2}{w^2+x^2}) \geqq 4$$

[DOTOANNANG]

$$\sum_{cyc}^{ \,\,\,\,\,\,\,\,\,\,\,}{\frac{x}{y+z}\,\,\sqrt{\,\frac{y+z}{7\,x+y+z}}\,}\geqq \frac{\sqrt{2}}{\,\,2}$$

[DOTOANNANG]

$$\left( {\frac {x}{x+y}}+{\frac {y}{z+x}} \right)\,\, \left( {\frac {y}{y +z}}+{\frac {z}{x+y}} \right)\,\, \left( {\frac {z}{z+x}}+{\frac {x}{y+z} } \right) \geqq 1$$

[DOTOANNANG]

$$\left( {\frac {y+z}{x}}+27\,{\frac {x}{x+y}} \right)\,\, \left( {\frac { z+x}{y}}+27\,{\frac {y}{y+z}} \right)\,\, \left( {\frac {x+y}{z}}+27\,{ \frac {z}{z+x}} \right) \geqq {\frac {29791}{8}}$$

[DOTOANNANG]

$$\sqrt {{\frac {x}{y+z}}}+\sqrt {{\frac {y}{z+x}}}+\sqrt {{\frac {z}{x+ y}}}>2$$

[DOTOANNANG]

$$\left ( \frac{x}{y+ z}+ \frac{9\,(y+ z)}{10\,x+ y+ z} \right )\,\,\left ( \frac{y}{z+ x}+ \frac{9\,(z+ x)}{10\,y+ z+ x} \right )\,\,\left ( \frac{z}{x+ y}+ \frac{9\,(x+ y)}{10\,z+ x+ y} \right )\geqq 8$$

[DOTOANNANG]

$$\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}+6\geqq 4\,\,(\frac{x+y}{y+z}+\frac{y+z}{z+x}+\frac{z+x}{x+y})$$

[DOTOANNANG]

$$x+ y+ z= 3$$

 

$$1+\frac{y}{z+x}\geqq \frac{3}{8}\,(x+y)\,(y+z)$$

[DOTOANNANG]

$$3+{\frac {x}{y+z}}+{\frac {y}{z+x}}+{\frac {z}{x+y}}\geqq \left [\, \left ( \frac{x}{y+ z} \right )^{1/\,2}+ \left ( \frac{y}{z+ x} \right )^{1/\,2}+ \left ( \frac{z}{x+ y} \right )^{1/\,2}\right ] ^{2}$$

[DOTOANNANG]

$$\frac{3}{8}\,\frac{(y+x)\,(2\,y+3\,z+x)}{(x+y+z)\,y}\geqq 2\,\frac{x}{y+x}+\frac{z}{y+z}$$

[DOTOANNANG]

[DOTOANNANG]

Cho các số $x\,,y\,,z$ không âm thỏa $yz + zx + xy > 0$, khi đó:

 

 

$$\frac {x}{y + z} + \frac {y}{z + x} + \frac {z}{x + y} \geqq 2 - \sqrt {2} + \left(\sqrt {2} - \frac {1}{2}\right)\frac { y^2z + z^2x + x^2y}{yz^2 + zx^2 + xy^2 }$$

 

 

Đẳng thức xảy ra khi và chỉ khi: $x = 4 + \sqrt {2},\,y = 3 - \sqrt {2},\,z = 0$

[DOTOANNANG]

$$\sqrt{\frac{x}{4\,y+z}}+\sqrt{\frac{y}{z+ 4\,x}}+\sqrt{\frac{z}{2\left ( x+ y \right )}}\geqq 1$$

[DOTOANNANG]

$$\frac{12\left \{ xy+ yz+ zx \right \}}{\left \{ x+ y+ z \right \}^{2}} + 2\left \{ \frac{x}{y+ z}+ \frac{y}{z+ x}+ \frac{z}{x+ y} \right \} \geqq 7$$

[DOTOANNANG]

$$\frac{\left \{ x+ y+ z \right \}^{2}}{xy+ yz+ zx}+ \frac{3\left \{ xy+ yz+ zx \right \}}{4\left \{ x+ y+ z \right \}^{2}}\geqq \frac{x}{y+ z}+ \frac{y}{z+ x}+ \frac{z}{x+ y}+ \frac{7}{4}$$

[DOTOANNANG]

$$\frac{\left \{ x+ y+ z \right \}^{2}}{xy+ yz+ zx}+ \frac{3\left \{ xy+ yz+ zx \right \}}{4\left \{ x+ y+ z \right \}^{2}}+ {\it \prod_{cycl}^{ }}\,\frac{x}{x+ y}{\it \prod_{cycl}^{ }}\,\frac{y}{x+ y}\geqq \frac{x}{y+ z}+ \frac{y}{z+ x}+ \frac{z}{x+ y}+ 4$$

[thanhdatqv2003]

                                                                           $\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{a+c}}+\frac{c}{\sqrt{a+b}}\geqslant \frac{1}{\sqrt{2}}(\sqrt{a}+\sqrt{b}+\sqrt{c})$

Bài viết đã được chỉnh sửa nội dung bởi thanhdatqv2003: 24-07-2018 - 23:52