Cho $ x,y\geq 0 và x+y=2.$ Tìm min, max của P=$\frac{x}{x+1}+\frac{2y}{y+2}$
$P=\left(1-\frac{1}{x+1}\right)+\left(2-\frac{4}{y+2}\right)=3-\left( \frac 1{x+1} +\frac 4{y+2} \right) \le 3-\frac{9}{x+1+y+2}=\frac 65$
Vậy $\max P=\frac 65$ đạt đc khi và chỉ khi $x=\frac 23,y=\frac 43$
Ta có $x=2-y \le 2$
Lại có $P=\frac x{x+1}+\frac{2(2-y)}{(2-y)+2}=\frac{x}{x+1}+\frac{4-2x}{4-x}=\frac{-3x^2+6x+4}{-x^2+3x+4}$
$P \ge \frac 23 \Leftrightarrow 3(-3x^2+6x+4) \ge 2(-x^2+3x+4)\Leftrightarrow (x-2)(7x+2) \le0$ (luôn đúng)
Vậy $\min P=\frac 23$ đạt được khi và chỉ khi $x=2,y=0$

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