Chủ đề này có 6 trả lời

[chibiwonder]

Cho x,y,x>0 ; x+y+z=1.a)CM: $\frac{yz}{1-x}+\frac{xz}{1-y}+\frac{yx}{1-z}\leqslant \frac{1}{2}$

b) $(\frac{1}{1+x})(\frac{1}{1+y})(\frac{1}{z+1})\geqslant 64$

c) $\sqrt{\frac{yz}{x+yz}}+\sqrt{\frac{zx}{zx+y}}+\sqrt{\frac{xy}{xy+z}}\leq \frac{3}{2}$

d) $\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y} \geqslant \frac{1}{2}$

[phamngochung9a]

Cho x,y,x>0 ; x+y+z=1.a)CM: $\frac{yz}{1-x}+\frac{xz}{1-y}+\frac{yx}{1-z}\leqslant \frac{1}{2}$

b) $(\frac{1}{1+x})(\frac{1}{1+y})(\frac{1}{z+1})\geqslant 64$

c) $\sqrt{\frac{yz}{x+yz}}+\sqrt{\frac{zx}{zx+y}}+\sqrt{\frac{xy}{xy+z}}\leq \frac{3}{2}$

d) $\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y} \geqslant \frac{1}{2}$

Làm câu dễ nhất 

d) Áp dụng BĐT Cauchy- schwarz:

 $\sum \frac{x^{2}}{y+z}\geq \frac{x+y+z}{2}=\frac{1}{2}$

[phamquanglam]

Cho x,y,x>0 ; x+y+z=1.a)CM: $\frac{yz}{1-x}+\frac{xz}{1-y}+\frac{yx}{1-z}\leqslant \frac{1}{2}$

 

ta thay $1-x=y+z$ ta có: $\frac{yz}{1-x}=\frac{yz}{y+z}\leq \frac{yz}{2\sqrt{yz}}=\frac{\sqrt{yz}}{2}\Rightarrow \sum \frac{yz}{1-x}\leq \sum \frac{\sqrt{yz}}{2}\leq \frac{x+y+z}{2}=\frac{1}{2}$

[phamngochung9a]

Cho x,y,x>0 ; x+y+z=1.a)CM: $\frac{yz}{1-x}+\frac{xz}{1-y}+\frac{yx}{1-z}\leqslant \frac{1}{2}$

b) $(\frac{1}{1+x})(\frac{1}{1+y})(\frac{1}{z+1})\geqslant 64$

c) $\sqrt{\frac{yz}{x+yz}}+\sqrt{\frac{zx}{zx+y}}+\sqrt{\frac{xy}{xy+z}}\leq \frac{3}{2}$

d) $\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y} \geqslant \frac{1}{2}$

a) $\sum \frac{yz}{1-x}=\sum \frac{yz}{y+z}\leq \frac{1}{4}(x+y+y+z+z+x)=\frac{1}{2}$

b) $\prod \frac{1}{1+x}=\frac{1}{(1+x)(1+y)(1+z)}\geq \frac{27}{(3+x+y+z)^{3}}=\frac{27}{64}$

c) $\sum \sqrt{\frac{yz}{x+yz}}=\sum \sqrt{\frac{yz}{(x+z)(x+y)}}\leq \frac{1}{2}\sum \left (\frac{z}{x+z}+\frac{y}{x+y} \right )=\frac{3}{2}$

[thuy99]

Cho x,y,x>0 ; x+y+z=1

a)CM: $\frac{yz}{1-x}+\frac{xz}{1-y}+\frac{yx}{1-z}\leqslant \frac{1}{2}$

 

đpcm $\Leftrightarrow \sum \frac{yz}{y+z}\leq \frac{x+y+z}{2}$

ta cm $\frac{yz}{y+z}\leq \frac{y+z}{4}\Leftrightarrow (y-z)^2\geq 0$ (luôn đúng) 

tương tự ta có dpcm

[quan1234]

Câu d, Áp dụng BĐT Cô-si,ta có

$\frac{x^2}{y+z}+\frac{y+z}{4}\geq x$

2 cái còn lại tương tự rồi cộng lại, ta được đpcm

Bài viết đã được chỉnh sửa nội dung bởi quan1234: 24-07-2015 - 21:03

[Dinh Xuan Hung]

Cho x,y,x>0 ; x+y+z=1.a)CM: $\frac{yz}{1-x}+\frac{xz}{1-y}+\frac{yx}{1-z}\leqslant \frac{1}{2}$

b) $(\frac{1}{1+x})(\frac{1}{1+y})(\frac{1}{z+1})\geqslant 64$

c) $\sqrt{\frac{yz}{x+yz}}+\sqrt{\frac{zx}{zx+y}}+\sqrt{\frac{xy}{xy+z}}\leq \frac{3}{2}$

d) $\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y} \geqslant \frac{1}{2}$

d)cách khác:

$AM-GM$:

$\frac{x^2}{y+z}+\frac{y+z}{4}\geq x$

$\frac{y^2}{x+z}+\frac{x+z}{4}\geq y$

$\frac{z^2}{x+y}+\frac{x+y}{4}\geq z$

$\Rightarrow \sum \frac{x^2}{y+z}+\frac{x+y+z}{2}\geq x+y+z\Leftrightarrow \sum \frac{x^2}{y+z}\geq \frac{x+y+z}{2}=\frac{1}{2}$