Áp dụng Cauchy-Schwarz thì $P\geq \frac{(x+y)^2}{2-(x+y)}+\frac{1}{x+y}+x+y=\frac{t^2}{2-t}+\frac{1}{t}+t=\frac{t^2}{2-t}+(\frac{1}{t}+\frac{9t}{4})-\frac{5t}{4}$ Với $00$$\Rightarrow P\geq \frac{4t^2-5t(2-t)}{4(2-t)}+2.\sqrt{\frac{9t}{4t}}=\frac{9t^2-10t}{4(2-t)}+3$Ta cm $\frac{9t^2-10t}{4(2-t)}\geq \frac{-1}{2} (*)$Thật vậy $(*)\Leftrightarrow \frac{9(t-\frac{2}{3})^2}{2(2-t)}\geq 0$ đúng do $2-t>0$ $\Rightarrow P\geq \frac{5}{2}$Dấu $"="$ xảy ra khi $x=y=\frac{1}{3}$Vậy $Min P=\frac{5}{2}$
Áp dụng Cauchy-Schwarz thì $P\geq \frac{(x+y)^2}{2-(x+y)}+\frac{1}{x+y}+x+y=\frac{t^2}{2-t}+\frac{1}{t}+t=\frac{t^2}{2-t}+(\frac{1}{t}+\frac{9t}{4})-\frac{5t}{4}$ Với $0<t=x+y<2\Rightarrow 2-t>0$$\Rightarrow P\geq \frac{4t^2-5t(2-t)}{4(2-t)}+2.\sqrt{\frac{9t}{4t}}=\frac{9t^2-10t}{4(2-t)}+3$Ta cm $\frac{9t^2-10t}{4(2-t)}\geq \frac{-1}{2} (*)$Thật vậy $(*)\Leftrightarrow \frac{9(t-\frac{2}{3})^2}{2(2-t)}\geq 0$ đúng do $2-t<0$ $\Rightarrow P\geq \frac{5}{2}$Dấu $"="$ xảy ra khi $x=y=\frac{1}{3}$Vậy $Min P=\frac{5}{2}$
Áp dụng Cauchy-Schwarz thì $P\geq \frac{(x+y)^2}{2-(x+y)}+\frac{1}{x+y}+x+y=\frac{t^2}{2-t}+\frac{1}{t}+t=\frac{t^2}{2-t}+(\frac{1}{t}+\frac{9t}{4})-\frac{5t}{4}$ Với $00$$\Rightarrow P\geq \frac{4t^2-5t(2-t)}{4(2-t)}+2.\sqrt{\frac{9t}{4t}}=\frac{9t^2-10t}{4(2-t)}+3$Ta cm $\frac{9t^2-10t}{4(2-t)}\geq \frac{-1}{2} (*)$Thật vậy $(*)\Leftrightarrow \frac{9(t-\frac{2}{3})^2}{2(2-t)}\geq 0$ đúng do $2-t&
gt;0$ $\Rightarrow P\geq \frac{5}{2}$Dấu $"="$ xảy ra khi $x=y=\frac{1}{3}$Vậy $Min P=\frac{5}{2}$