Bai toan cuc tri

C

chonhoi110

Ta có $x \ge 2 \Longrightarrow y \ge 3-x \ge 1$

$\Longrightarrow (x-2)^2+(y-1)^2 \ge 0 \Longrightarrow x^2+y^2\ge 4x+2y-5$

$\Longrightarrow A=x^2+y^2+\dfrac{1}{x}+\dfrac{1}{x+y}\ge 4x+2y-5 +\dfrac{1}{x}+\dfrac{1}{x+y}$

$=(\dfrac{x}{4}+ \dfrac{1}{x})+(\dfrac{x+y}{9}+ \dfrac{1}{x+y})+ \dfrac{17(x+y)}{9}+ \dfrac{7x}{4}-5$

$\ge 2\sqrt{\dfrac{x}{4}.\dfrac{1}{x}} +2\sqrt{\dfrac{x+y}{9}.\dfrac{1}{x+y}}+\dfrac{17(x+y)}{9}+\dfrac{7x}{4}-5 \ge 1+\dfrac{2}{3}+\dfrac{17}{3}+\dfrac{7}{2}-5=\dfrac{35}{6}$

Vậy Min $A=.... \Longleftrightarrow x=2; y=1$
 
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