Chủ đề này có 2 trả lời

[DOTOANNANG]

Cho $a,\,b,\,c>0,\,abc= 1$. Chứng minh:

 

$$\frac{1}{1+ \sqrt{ab+ 3\,c}}+ \frac{1}{1+ \sqrt{bc+ 3\,a}}+ \frac{1}{1+ \sqrt{ca+ 3\,b}}\leqq 1$$

[Arthur Pendragon]

Giải:

Đặt $a=\frac{x}{y}....$

Bđt cần cm tương đương với:

$\frac{\sqrt{xz}}{\sqrt{xz}+\sqrt{x^2+3z^2}}+\frac{\sqrt{zy}}{\sqrt{zy}+\sqrt{z^2+3y^2}}+\frac{\sqrt{yx}}{\sqrt{yx}+\sqrt{y^2+3x^2}} \leq 1$

Ta có:

$\frac{\sqrt{xz}}{\sqrt{xz}+\sqrt{x^2+3z^2}}+\frac{\sqrt{zy}}{\sqrt{zy}+\sqrt{z^2+3y^2}}+\frac{\sqrt{yx}}{\sqrt{yx}+\sqrt{y^2+3x^2}} $

$\leq \frac{\sqrt{xz}}{\sqrt{xz}+\sqrt{2xz+2z^2}}+\frac{\sqrt{zy}}{\sqrt{zy}+\sqrt{2zy+2y^2}}+\frac{\sqrt{yx}}{\sqrt{yx}+\sqrt{2yx+2x^2}}$

$\leq \frac{\sqrt{xz}}{\sqrt{xz}+\sqrt{2z(x+z)}}+\frac{\sqrt{zy}}{\sqrt{zy}+\sqrt{2y(z+y)}}+\frac{\sqrt{yx}}{\sqrt{yx}+\sqrt{2x(y+x)}}$

$=\frac{\sqrt{x}}{\sqrt{x}+\sqrt{2(x+z)}+\frac{\sqrt{z}}{\sqrt{z}+\sqrt{2(y+z)}+\frac{\sqrt{y}}{\sqrt{y}+\sqrt{2(y+x)}$

$\leq \frac{\sqrt{x}}{\sqrt{x}+\sqrt{x}+\sqrt{z}}+\frac{\sqrt{z}}{\sqrt{z}+\sqrt{y}+\sqrt{z}}+\frac{\sqrt{y}}{\sqrt{y}+\sqrt{y}+\sqrt{x}}$

Ta cần cm $\frac{\sqrt{x}}{2\sqrt{x}+\sqrt{z}}+\frac{\sqrt{z}}{2\sqrt{z}+\sqrt{y}}+\frac{\sqrt{y}}{2\sqrt{y}+\sqrt{x}} \leq 1$

 

điều này tương đương với $\frac{\sqrt{z}}{2\sqrt{x}+\sqrt{z}}+\frac{\sqrt{y}}{2\sqrt{z}+\sqrt{y}}+\frac{\sqrt{x}}{2\sqrt{y}+\sqrt{x}} \geq 1$

Ez game...

[DOTOANNANG]

$2\,.$ Với $a,\,b,\,c> 0$ & $a^{2}+ b^{2}+ c^{2}= 3$ thì:

 

$$\frac{1}{1+ \sqrt{ab+ 3\,c}}+ \frac{1}{1+ \sqrt{bc+ 3\,a}}+ \frac{1}{1+ \sqrt{ca+ 3\,b}}\geqq 1$$

 

Spoiler

 

Đặt $x,\,y,\,z= \sqrt{ab+ 3\,c},\,\sqrt{bc+ 3\,a},\,\sqrt{ca+ 3\,b}$

 

$x^{2}+ y^{2}+ z^{2}= \underbrace{\sum\limits_{cyc}\left ( ab+ 3\,c \right )\leqq 12  }_{a^{2}+ b^{2}+ c^{2}= 3}$

 

$\Leftrightarrow \sum\limits_{cyc}\frac{1}{1+ x}\geqq \frac{9}{3+ \sum\limits_{cyc}x }\geqq \frac{9}{3+ \sqrt{3\sum\limits_{cyc}x^{2} }}\geqq 1$