2 replies to this topic

[githenhi512]

Cho x,y>1 t/m: x+y+3=xy. Tìm Max P= $\frac{\sqrt{x^2-1}}{x}+\frac{\sqrt{y^2-1}}{y}+\frac{1}{x+y}$

[leminhnghiatt]

Cho x,y>1 t/m: x+y+3=xy. Tìm Max P= $\frac{\sqrt{x^2-1}}{x}+\frac{\sqrt{y^2-1}}{y}+\frac{1}{x+y}$

 

$x+y+3=xy \leq \dfrac{(x+y)^2}{4} \rightarrow (x+y)^2-4(x+y)-12 \geq 0 \rightarrow (x+y-6)(x+y+2) \geq 0 \rightarrow x+y \geq 6$

 

TT: $x+y+3=xy \rightarrow \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{3}{xy}=1$

 

$\rightarrow 1 \leq \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{3}{4}(\dfrac{1}{x}+\dfrac{1}{y})^2 \rightarrow \dfrac{1}{x}+\dfrac{1}{y} \geq \dfrac{2}{3}$

 

Ta có: $P=\sqrt{1-\dfrac{1}{x^2}}+\sqrt{1-\dfrac{1}{y^2}}+\dfrac{1}{x+y}$

 

$= \dfrac{3\sqrt{\dfrac{8}{9}(1-\dfrac{1}{x^2})}}{2\sqrt{2}}+\dfrac{3\sqrt{\dfrac{8}{9}(1-\dfrac{1}{y^2})}}{2\sqrt{2}}+\dfrac{1}{x+y}$

 

$\leq \dfrac{3(\dfrac{8}{9}+1-\dfrac{1}{x^2})}{4\sqrt{2}}+\dfrac{3(\dfrac{8}{9}+1-\dfrac{1}{y^2})}{4\sqrt{2}}+\dfrac{1}{x+y}$

 

$= \dfrac{17\sqrt{2}}{12}-\dfrac{3(\dfrac{1}{x^2}+\dfrac{1}{y^2})}{4\sqrt{2}}+\dfrac{1}{6}$

 

$\leq \dfrac{17\sqrt{2}}{12}-\dfrac{3(\dfrac{1}{x}+\dfrac{1}{y})^2)}{8\sqrt{2}}+\dfrac{1}{6}$

 

$\leq  \dfrac{17\sqrt{2}}{12}-\dfrac{3.(\dfrac{2}{3})^2}{8\sqrt{2}}+\dfrac{1}{6}$

 

$=\dfrac{1+8\sqrt{2}}{6}$

 

Dấu "=" $\iff x=y=3$

Edited by leminhnghiatt, 30-05-2016 - 09:42.

[Thislife]

Cho x,y>1 t/m: x+y+3=xy. Tìm Max P= $\frac{\sqrt{x^2-1}}{x}+\frac{\sqrt{y^2-1}}{y}+\frac{1}{x+y}$

Cách khác $:$

Ta có P=$\sqrt{1-\frac{1}{x^{2}}} +\sqrt{1-\frac{1}{y^{2}}} +\frac{1}{x+y}\leq \sqrt{2(2-\frac{1}{x^{2}}-\frac{1}{y^{2}})} +\frac{1}{4}*(\frac{1}{x}+\frac{1}{y})$ ( $Cauchy-Schwarz$ )

$\Rightarrow P \leq \sqrt{2(2-\frac{(\frac{1}{x}+\frac{1}{y})^{2}}{2})}+ \frac{1}{4}*(\frac{1}{x}+\frac{1}{y}) $

Từ điều kiện x,y>1 $\Rightarrow \frac{1}{x}+\frac{1}{y} < 2$

và $$ x+y+3=xy \Rightarrow 3=\frac{1}{x} +\frac{1}{y} +\frac{3}{xy} \leq \frac{1}{x} +\frac{1}{y} +\frac{3}{4}*(\frac{1}{x} +\frac{1}{y})^{2}(AM-GM)$$

Đặt $\frac{1}{x} +\frac{1}{y} =a$  Ta được :

$$\begin{cases} 3\leq a+\frac{3}{4}*a^{2} \\a<2 \end{cases}$$

$$\Rightarrow \frac{2}{3} \leq a  <2 $$

Xét $P=f(a)=\sqrt{4-a^{2}} +\frac{1}{4}a$  trên $ [\frac{2}{3};2)$ ta được $f(a)$ nghịch biến 

Nên $\Rightarrow f(a) \leq f(\frac{2}{3}) = \frac{1+8\sqrt{2}}{6}$