Tìm Max

M

minhtuyb

Ta chỉ xét trường hợp x,y>0x,y>0
Từ gt ta có: xxy+12xyx4yxy4x\ge xy+1\ge 2\sqrt{xy}\Rightarrow x\ge 4y\Leftrightarrow \dfrac{x}{y}\ge 4
PP max 1P\Leftrightarrow \dfrac{1}{P} min:
1P=x3y+y3x=(x48y+y3x)+516.xy2x48y.y3x+516.4=1217P1712 (const)\dfrac{1}{P}=\dfrac{x}{3y}+\dfrac{y}{3x} \\ =(\dfrac{x}{48y}+\dfrac{y}{3x})+\dfrac{5}{16}.\dfrac{x}{y}\\ \ge 2\sqrt{\dfrac{x}{48y}.\dfrac{y}{3x}}+\dfrac{5}{16}.4\\ =\dfrac{12}{17}\\ \Rightarrow P\le \dfrac{17}{12}\ (const)
Dấu bằng xảy ra khi x=2;y=12 x=2;y=\dfrac{1}{2}\ \square
 
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