Bài này số xấu làm lằng nhằng qáBài 468: Cho 3 số dương x,y,z có x+y+z=1. Chứng minh:
$\frac{1+\sqrt{x}}{y+z}+\frac{1+\sqrt{y}}{z+x}+\frac{1+\sqrt{z}}{x+y}\geq \frac{9+3\sqrt{3}}{2}$
Ta có : $\frac{1+\sqrt{x}}{y+z}= \frac{1+\sqrt{x}}{1-x}= \frac{1}{1-\sqrt{x}}= \frac{1}{1-\frac{x}{2\sqrt{x.\frac{1}{3}}}.\frac{2}{\sqrt{3}}}\geq \frac{1}{1-\frac{x}{x+\frac{1}{3}}.\frac{2}{\sqrt{3}}}$
$= \frac{3x+1}{3x+1-2\sqrt{3}.x}$
Vậy ta chỉ cần CM :
$\sum \frac{3x+1}{3x+1-2\sqrt{3}.x}\geq \frac{9+3\sqrt{3}}{2}\Leftrightarrow \sum \frac{3x}{3x+1-2\sqrt{3}.x}+\sum \frac{1}{3x+1-2\sqrt{3}.x}\geq \frac{9+3\sqrt{3}}{2}$
Ta có :
$\sum \frac{3x}{3x+1-2\sqrt{3}.x}= \sum \frac{3x^{2}}{3x^{2}+x-2\sqrt{3}x^{2}}\geq \frac{3(x+y+z)^{2}}{(x+y+z)-(2\sqrt{3}-3)(x^{2}+y^2+z^2)}$
$\geq \frac{3}{1-(2\sqrt{3}-3)\frac{(x+y+z)^{2}}{3}}= \frac{3}{1-(2\sqrt{3}-3).\frac{1}{3}}= \frac{9}{6-2\sqrt{3}}$
Và :
$\sum \frac{1}{3x+1-2\sqrt{3}.x}\geq \frac{9}{3(x+y+z)+3-2\sqrt{3}(x+y+z)}= \frac{9}{6-2\sqrt{3}}$
Vậy ;
$VT\geq \frac{9}{6-2\sqrt{3}}+\frac{9}{6-2\sqrt{3}}= \frac{9}{3-\sqrt{3}}= \frac{9+3\sqrt{3}}{2}$
( trục căn thúc )