Chủ đề này có 3 trả lời

[yeutoanmaimai1]

Cho $a,b,c>0$ thỏa mãn $abc=1$ Chứng minh $\frac{1}{(1+a)^{2}+b^{2}+1}+\frac{1}{(1+b)^{2}+c^{2}+1}+\frac{1}{(1+c)^{2}+a^{2}+1}\leq \frac{1}{2}$

2,Cho $x,y,z>0$ thỏa mãn $xy+yz+xz=670$ Chứng minh $\frac{x}{x^{2}-yz+2010}+\frac{y}{y^{2}-xz+2010}+\frac{z}{z^{2}-xy+2010}\geq \frac{1}{x+y+z}$

Bài viết đã được chỉnh sửa nội dung bởi yeutoanmaimai1: 18-03-2015 - 11:05

[Dinh Xuan Hung]

Cho $a,b,c>0$ thỏa mãn $abc=1$ Chứng minh $\frac{1}{(1+a)^{2}+b^{2}+1}+\frac{1}{(1+b)^{2}+c^{2}+1}+\frac{1}{(1+c)^{2}+a^{2}+1}\leq \frac{1}{2}$

2,Cho $x,y,z>0$ thỏa mãn $xy+yz+xz=670$ Chứng minh $\frac{x}{x^{2}-yz+2010}+\frac{y}{y^{2}-xz+2010}+\frac{z}{z^{2}-xy+2010}\geq \frac{1}{x+y+z}$

1.

$\sum \frac{1}{(a+1)^2+b^2+1}=\sum \frac{1}{a^2+b^2+2a+2}\leq \sum \frac{1}{2ab+2a+2}=\frac{1}{2}\left ( \frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ac+c+1} \right )=\frac{1}{2}$

[Dinh Xuan Hung]

 

2,Cho $x,y,z>0$ thỏa mãn $xy+yz+xz=670$ Chứng minh $\frac{x}{x^{2}-yz+2010}+\frac{y}{y^{2}-xz+2010}+\frac{z}{z^{2}-xy+2010}\geq \frac{1}{x+y+z}$

Dễ có $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$

$\sum \frac{x}{x^2-yz+2010}=\sum \frac{x^2}{x^3-xyz+2010x}\geq \frac{(x+y+z)^2}{x^3+y^3+z^3-3xyz+2010(x+y+z)}=\frac{(x+y+z)^2}{(x+y+z)(x^2+y^2+z^2-xy-yz-xz)+2010(x+y+z)}=\frac{(x+y+z)^2}{(x+y+z)(x^2+y^2+z^2+2xy+2yz+2xz-3(xy+yz+xz)+2010)}=\frac{(x+y+z)^2}{(x+y+z)^3}=\frac{1}{x+y+z}$

[Dinh Xuan Hung]

1.

$\sum \frac{1}{(a+1)^2+b^2+1}=\sum \frac{1}{a^2+b^2+2a+2}\leq \sum \frac{1}{2ab+2a+2}=\frac{1}{2}\left ( \frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ac+c+1} \right )=\frac{1}{2}$

$\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ac+c+1}=\frac{1}{ab+a+1}+\frac{a}{abc+ab+a}+\frac{abc}{ac+c+abc}=\frac{1}{ab+a+1}+\frac{a}{1+ab+a}+\frac{ab}{a+1+ab}=1$