Cho x,y>0 thỏa mãn x+y ≤ 1 Tìm GTNN của B=$\frac{1}{x+y}$ ² ² +$\frac{2}{xy}$

18/08/2021 Bởi Allison

Cho x,y>0 thỏa mãn x+y ≤ 1 Tìm GTNN của
B=$\frac{1}{x+y}$ ² ² +$\frac{2}{xy}$

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thucquyen

18/08/2021 vào lúc 10:31

Giải thích các bước giải:

$x + y = 1 \Rightarrow x = 1 - y$ $x + y = 1 \Rightarrow x = 1 - y$

$\Rightarrow x^{2} + y^{2} = {(1 - y)}^{2} + y^{2} = 2 y^{2} - 2 y + 1 = 2 (y^{2} - y + \frac{1}{2}) = 2 (y^{2} - 2 y \cdot \frac{1}{2} + \frac{1}{4}) + \frac{1}{2} = 2 {(y - \frac{1}{2})}^{2} + \frac{1}{2} \geq \frac{1}{2}$

$\Rightarrow x^{2} + y^{2} = {(1 - y)}^{2} + y^{2} = 2 y^{2} - 2 y + 1 = 2 (y^{2} - y + \frac{1}{2}) = 2 (y^{2} - 2 y \cdot \frac{1}{2} + \frac{1}{4}) + \frac{1}{2} = 2 {(y - \frac{1}{2})}^{2} + \frac{1}{2} \geq \frac{1}{2}$

$\Rightarrow x^{2} + y^{2} = {(1 - y)}^{2} + y^{2} = 2 y^{2} - 2 y + 1 = 2 (y^{2} - y + \frac{1}{2}) = 2 (y^{2} - 2 y \cdot \frac{1}{2} + \frac{1}{4}) + \frac{1}{2} = 2 {(y - \frac{1}{2})}^{2} + \frac{1}{2} \geq \frac{1}{2}$

$\Rightarrow x^{2} + y^{2} = {(1 - y)}^{2} + y^{2} = 2 y^{2} - 2 y + 1 = 2 (y^{2} - y + \frac{1}{2}) = 2 (y^{2} - 2 y \cdot \frac{1}{2} + \frac{1}{4}) + \frac{1}{2} = 2 {(y - \frac{1}{2})}^{2} + \frac{1}{2} \geq \frac{1}{2}$

$\Rightarrow x^{2} + y^{2} = {(1 - y)}^{2} + y^{2} = 2 y^{2} - 2 y + 1 = 2 (y^{2} - y + \frac{1}{2}) = 2 (y^{2} - 2 y \cdot \frac{1}{2} + \frac{1}{4}) + \frac{1}{2} = 2 {(y - \frac{1}{2})}^{2} + \frac{1}{2} \geq \frac{1}{2}$

$\Rightarrow x^{2} + y^{2} = {(1 - y)}^{2} + y^{2} = 2 y^{2} - 2 y + 1 = 2 (y^{2} - y + \frac{1}{2}) = 2 (y^{2} - 2 y \cdot \frac{1}{2} + \frac{1}{4}) + \frac{1}{2} = 2 {(y - \frac{1}{2})}^{2} + \frac{1}{2} \geq \frac{1}{2}$

$\Rightarrow x^{2} + y^{2} = {(1 - y)}^{2} + y^{2} = 2 y^{2} - 2 y + 1 = 2 (y^{2} - y + \frac{1}{2}) = 2 (y^{2} - 2 y \cdot \frac{1}{2} + \frac{1}{4}) + \frac{1}{2} = 2 {(y - \frac{1}{2})}^{2} + \frac{1}{2} \geq \frac{1}{2}$

$\Rightarrow x^{2} + y^{2} = {(1 - y)}^{2} + y^{2} = 2 y^{2} - 2 y + 1 = 2 (y^{2} - y + \frac{1}{2}) = 2 (y^{2} - 2 y \cdot \frac{1}{2} + \frac{1}{4}) + \frac{1}{2} = 2 {(y - \frac{1}{2})}^{2} + \frac{1}{2} \geq \frac{1}{2}$

Vậy $A_{M i n} = \frac{1}{2} \Leftrightarrow x = y = \frac{1}{2}$

2.

Ta có:

$B = \frac{1}{x^{2} y^{2}} - \frac{1}{x^{2}} - \frac{1}{y^{2}} = \frac{1}{x^{2} y^{2}} - \frac{y^{2}}{x^{2} y^{2}} - \frac{x^{2}}{x^{2} y^{2}} = \frac{1 - (x^{2} + y^{2})}{x^{2} y^{2}} \leq \frac{1 - \frac{1}{2}}{\frac{1}{4} \cdot \frac{1}{4}} = \frac{\frac{1}{2}}{\frac{1}{8}} = \frac{1}{4}$

$B = \frac{1}{x^{2} y^{2}} - \frac{1}{x^{2}} - \frac{1}{y^{2}} = \frac{1}{x^{2} y^{2}} - \frac{y^{2}}{x^{2} y^{2}} - \frac{x^{2}}{x^{2} y^{2}} = \frac{1 - (x^{2} + y^{2})}{x^{2} y^{2}} \leq \frac{1 - \frac{1}{2}}{\frac{1}{4} \cdot \frac{1}{4}} = \frac{\frac{1}{2}}{\frac{1}{8}} = \frac{1}{4}$

$B = \frac{1}{x^{2} y^{2}} - \frac{1}{x^{2}} - \frac{1}{y^{2}} = \frac{1}{x^{2} y^{2}} - \frac{y^{2}}{x^{2} y^{2}} - \frac{x^{2}}{x^{2} y^{2}} = \frac{1 - (x^{2} + y^{2})}{x^{2} y^{2}} \leq \frac{1 - \frac{1}{2}}{\frac{1}{4} \cdot \frac{1}{4}} = \frac{\frac{1}{2}}{\frac{1}{8}} = \frac{1}{4}$

$B = \frac{1}{x^{2} y^{2}} - \frac{1}{x^{2}} - \frac{1}{y^{2}} = \frac{1}{x^{2} y^{2}} - \frac{y^{2}}{x^{2} y^{2}} - \frac{x^{2}}{x^{2} y^{2}} = \frac{1 - (x^{2} + y^{2})}{x^{2} y^{2}} \leq \frac{1 - \frac{1}{2}}{\frac{1}{4} \cdot \frac{1}{4}} = \frac{\frac{1}{2}}{\frac{1}{8}} = \frac{1}{4}$

$B = \frac{1}{x^{2} y^{2}} - \frac{1}{x^{2}} - \frac{1}{y^{2}} = \frac{1}{x^{2} y^{2}} - \frac{y^{2}}{x^{2} y^{2}} - \frac{x^{2}}{x^{2} y^{2}} = \frac{1 - (x^{2} + y^{2})}{x^{2} y^{2}} \leq \frac{1 - \frac{1}{2}}{\frac{1}{4} \cdot \frac{1}{4}} = \frac{\frac{1}{2}}{\frac{1}{8}} = \frac{1}{4}$

$B = \frac{1}{x^{2} y^{2}} - \frac{1}{x^{2}} - \frac{1}{y^{2}} = \frac{1}{x^{2} y^{2}} - \frac{y^{2}}{x^{2} y^{2}} - \frac{x^{2}}{x^{2} y^{2}} = \frac{1 - (x^{2} + y^{2})}{x^{2} y^{2}} \leq \frac{1 - \frac{1}{2}}{\frac{1}{4} \cdot \frac{1}{4}} = \frac{\frac{1}{2}}{\frac{1}{8}} = \frac{1}{4}$

Vậy $B_{M a x} = \frac{1}{4} \Leftrightarrow x = y = \frac{1}{2}$

Chúc bn học tốt !

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