3 replies to this topic

[tienducfchn]

Cho a,b,c>0 thoa man ab+bc+ca+2abc=1. Chung minh $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq \frac{3}{2}$

[phuocdinh1999]

$GT\Leftrightarrow \sum \frac{1}{1+a}=2=\sum \frac{y+z}{x+y+z}=\sum \frac{1}{1+\frac{x}{y+z}}\: (x;y;z>0)$

Nên ta có thể đặt $a=\frac{x}{y+z};b=\frac{y}{x+z};c=\frac{z}{x+y}$

$\sum \sqrt{ab}=\sum \sqrt{\frac{xy}{(y+z)(x+z)}}\leq \sum \frac{1}{2}.(\frac{x}{x+z}+\frac{y}{y+z})=\frac{1}{2}.3=\frac{3}{2}$

[rohupt]

Một cách khác này:
Đặt $\sqrt{ab}=x; \sqrt{bc}=y; \sqrt{ca}=z; s=x+y+z$
Từ gt ta có $x^{2}+y^{2}+z^{2}+2xyz=1$
Xét
$s^{2}-2s+1$
$=(x+y+z)^{2}-2(x+y+z)+1$
$=1-2xyz+2(xy+yz+zx)-2(x+y+z)+1$
$=2(1-x)(1-y)(1-z) \leq 2(\frac{1-x+1-y+1-z}{3})^{3}$
$\Rightarrow s^{2}-2s+1 \leq 2(\frac{3-s}{3})^{3}$
$\Leftrightarrow 27(s^{2}-2s+1) \leq 2(27-27s+9s^{2}-s^{3})$
$\Leftrightarrow 2s^{3}+9s^{2}-27 \leq 0$
$\Leftrightarrow (2s-3)(s+3)^{2} \leq 0$
$\Leftrightarrow 2s-3 \leq 0 \Leftrightarrow s \leq \frac{3}{2} đpcm$

Đẳng thức $\Leftrightarrow 1-x=1-y=1-z \Leftrightarrow x=y=z \Leftrightarrow ab=bc=ca\Leftrightarrow a=b=c=\frac{1}{2}$

Edited by rohupt, 05-12-2013 - 21:29.

[tienducfchn]

 rohupt

 

vãi hiếu với mai

Edited by tienducfchn, 05-12-2013 - 21:24.