Chủ đề này có 6 trả lời

[DOTOANNANG]

$x\,, y\,, z\, > \, 0$. Chứng minh rằng:

\[{\frac {{x}^{\,2}}{{x}^{\,2}\,+\,{y}^{\,2}\,-\,xy}}\,+\,{\frac {{y}^{\,2}}{{y}^{\,2}\,+\,{z}^{\,2}\,-\,yz}}\,+\,{\frac {{z}^{\,2}}{{z}^{\,2}\,+\,{x}^{\,2}\,-\,xz\,}}\,\leq \,3\]

[Sauron]

$\sum\frac{x^2}{x^2+y^2-xy} = \sum\frac{x^2(x+y)}{x^3+y^3} = \sum\frac{x^3+x^2y}{x^3+y^3}= 3+\sum\frac{x^2y-y^3}{x^3+y^3} \leq 3+\sum\frac{y(x-y)(x+y)}{xy(x+y)}=3+\sum\frac{x-y}{x}=6-\frac{y}{x}-\frac{z}{y}-\frac{x}{z}\leq 3$.

[DOTOANNANG]

\[{\frac {{\,x}^{\,2}}{{\,x}^{\,2}\,+\,{\,y}^{\,2}\,+\,yz}}+{\frac {{\,y}^{\,2}}{{\,y}^{\,2}\,+\,{\,z}^{\,2}\,+\,xz}}\,+\,{\frac {{\,z}^{\,2}}{{\,z}^{\,2}\,+\,{\,x}^{\,2}\,+\,xy}}\,\geq\, 1\]

[DOTOANNANG]

\[\frac{\,3}{\,2}\,\geq \, \frac{\,y^{\,2}\,+\, 1\,}{\,2\,+\, xy\,+ \,x^{\,2}}\,+\, \frac{\,z^{\,2}\,+\, 1}{\,2\,+\, yz\,+\, y^{\,2}\,}+ \frac{\,x^{\,2}\,+ \,1\,}{\,2\,+\, xz\,+ \,z^{\,2}\,}\]

[DOTOANNANG]

\[\frac{\,x}{\,\sqrt{\,3\,x\,+\,2\,y\,+\,z\,}}\,+\,\frac{\,y}{\,\sqrt{\,3\,y\,+\,2\,z\,+\,x\,}}\,+\,\frac{\,z}{\,\sqrt{\,3\,z\,+\,2\,x\,+\,y\,}}\,\leq\,\sqrt{\,\frac{\,x\,+\,y\,+z\,}{\,2}}\]

[tr2512]

\[{\frac {{\,x}^{\,2}}{{\,x}^{\,2}\,+\,{\,y}^{\,2}\,+\,yz}}+{\frac {{\,y}^{\,2}}{{\,y}^{\,2}\,+\,{\,z}^{\,2}\,+\,xz}}\,+\,{\frac {{\,z}^{\,2}}{{\,z}^{\,2}\,+\,{\,x}^{\,2}\,+\,xy}}\,\geq\, 1\]

Áp dụng bất đẳng thức C-S:

$\sum \frac{{{x^2}}}{{{x^2} + {y^2} + yz}} \ge \frac{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^2}}}{{\sum {x^4} + \sum {x^2}{y^2} + xyz\sum x}} \ge \frac{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^2}}}{{\sum {x^4} + \sum {x^2}{y^2} + \sum {x^2}{y^2}}} = \frac{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^2}}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^2}}} = 1$

[Khoa Linh]

$\sum\frac{x^2}{x^2+y^2-xy} = \sum\frac{x^2(x+y)}{x^3+y^3} = \sum\frac{x^3+x^2y}{x^3+y^3}= 3+\sum\frac{x^2y-y^3}{x^3+y^3} \leq 3+\sum\frac{y(x-y)(x+y)}{xy(x+y)}=3+\sum\frac{x-y}{x}=6-\frac{y}{x}-\frac{z}{y}-\frac{x}{z}\leq 3$.

BĐT chỗ này bị sai:

$\sum\frac{x^2y-y^3}{x^3+y^3} \leq \sum\frac{y(x-y)(x+y)}{xy(x+y)}$

Mặc dù $x^3+y^3\geq xy(x+y)$ nhưng trên tử số là $y(x-y)(x+y)$ chưa là số dương nên bạn không thể làm thế được