3 replies to this topic

[Capture]

Cho x, y, z >0 . CMR:

$\sqrt{(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})}\geq 1+\sqrt{1+\sqrt{(x^{2}+y^{2}+z^{2})(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}})}}$

[Thao Huyen]

Cho x, y, z >0 . CMR:

$\sqrt{(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})}\geq 1+\sqrt{1+\sqrt{(x^{2}+y^{2}+z^{2})(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}})}}$

$a=\sum \frac{x}{y};b=\sum \frac{y}{x}\Rightarrow \sqrt{a+b+3}\geqslant 1+\sqrt{1+\sqrt{a^2+b^2-2a-2b+3}}$

Bình phương lên :v

[Lee LOng]

Ta có: 

$(\sum x)(\sum \frac{1}{x})=\sqrt{(\sum x^{2}+2\sum xy)(\sum \frac{1}{x^{2}}+2\sum \frac{1}{xy})}\geq \sqrt{\sum (x^{2})(\sum \frac{1}{x^{2}})}+2\sqrt{(\sum xy)(\sum \frac{1}{xy})} (*)$

BĐT $\Leftrightarrow (\sqrt{(\sum x)(\sum \frac{1}{x})}-1)^{2}\geq 1+\sqrt{(\sum x^{2})(\sum \frac{1}{x^{2}})}$ (Đúng theo $(*)$)

Dấu $(=) \Leftrightarrow (x^{2}-yz)(y^{2}-xz)(z^{2}-xy)=0$

Edited by Lee LOng, 26-07-2015 - 10:19.

[tranductucr1]

Ta có: 

$(\sum x)(\sum \frac{1}{x})=\sqrt{(\sum x^{2}+2\sum xy)(\sum \frac{1}{x^{2}}+2\sum \frac{1}{xy})}\geq \sqrt{\sum (x^{2})(\sum \frac{1}{x^{2}})}+2\sqrt{(\sum xy)(\sum \frac{1}{xy})} (*)$

BĐT $\Leftrightarrow (\sqrt{(\sum x)(\sum \frac{1}{x})}-1)^{2}\geq 1+\sqrt{(\sum x^{2})(\sum \frac{1}{x^{2}})}$ (Đúng theo $(*)$)

theo bạn dấu = xảy ra khi nào ? 

Edited by tranductucr1, 26-07-2015 - 09:40.